Explanation:
Okay, so I need to find the value of k for the ellipse given by the equation kx² + y² = 1, where k > 1. The problem states that there's a circle passing through the foci F₁ and F₂ of this ellipse and tangent to the ellipse at two points on the x-axis. Hmm, let me try to break this down step by step.

First, let me recall some properties of ellipses. The standard form of an ellipse is (x²/a²) + (y²/b²) = 1, where a and b are the semi-major and semi-minor axes. However, in this case, the equation is given as kx² + y² = 1. Let me rewrite this in standard form. If I divide both sides by 1, it's already in standard form, but I need to identify a² and b². Let me rearrange:

kx² + y² = 1 ⇒ (x²)/(1/k) + (y²)/1 = 1.

So comparing with the standard form, that would mean that a² = 1/k and b² = 1? Wait, but hold on. Wait, in the standard form, if the major axis is along the x-axis, then a² > b². However, here, since k > 1, then 1/k < 1. Therefore, the major axis is actually along the y-axis? Wait, no. Wait, if the coefficient of x² is k, which is greater than 1, then actually, if we write it as x²/(1/k) + y²/1 = 1, then since 1/k is less than 1, the denominator under x² is smaller, so the major axis is along the y-axis. Wait, but that would mean that the major axis is along the y-axis. However, the problem mentions that the circle is tangent to the ellipse at two points on the x-axis. So maybe the major axis is along the y-axis? Let me confirm.

Alternatively, maybe I should think of the ellipse as (x²)/(1/k) + y²/1 = 1. So if k > 1, then 1/k < 1, so the major axis is along the y-axis. Therefore, the foci are located along the y-axis. Wait, but the problem says that there is a circle passing through F₁ and F₂ and tangent to the ellipse at two points on the x-axis. If the foci are on the y-axis, then the circle passes through two points on the y-axis and is tangent to the ellipse at two points on the x-axis. That seems possible.

But let me check again. Let me recall that for an ellipse, the distance from the center to each focus is c, where c² = a² - b². Wait, but depending on whether the major axis is along x or y. Let me get this straight.

If the standard form is (x²/a²) + (y²/b²) = 1, then if a > b, the major axis is along the x-axis, and c² = a² - b². If b > a, then the major axis is along the y-axis, and c² = b² - a². So in our case, the equation is kx² + y² = 1. Let me write it as x²/(1/k) + y²/1 = 1. Therefore, if we let a² = 1/k and b² = 1, then if k > 1, then 1/k < 1, so b² > a², so the major axis is along the y-axis. Therefore, the foci are located along the y-axis at (0, ±c), where c² = b² - a² = 1 - 1/k. Therefore, c = sqrt(1 - 1/k). Therefore, the foci are at (0, sqrt(1 - 1/k)) and (0, -sqrt(1 - 1/k)). Wait, but the problem says that the circle passes through F₁ and F₂. So the circle passes through these two points on the y-axis. Also, the circle is tangent to the ellipse at two points on the x-axis. So the points of tangency are on the x-axis, meaning their y-coordinates are zero. Let me note that.

So, first, let me note the coordinates of the foci. Let me compute c. Since c² = b² - a² = 1 - 1/k. Therefore, c = sqrt(1 - 1/k). Therefore, the foci are at (0, c) and (0, -c). Therefore, the circle passes through (0, c) and (0, -c). Also, the circle is tangent to the ellipse at two points on the x-axis. Let me denote those points as (p, 0) and (-p, 0), since they are on the x-axis and symmetric with respect to the y-axis. Wait, but the ellipse equation is kx² + y² = 1. If we plug in y = 0, we get kx² = 1 ⇒ x² = 1/k ⇒ x = ±1/√k. Therefore, the ellipse intersects the x-axis at (±1/√k, 0). Therefore, if the circle is tangent to the ellipse at two points on the x-axis, those points must be (±1/√k, 0). Wait, but is that necessarily true? Wait, the circle is tangent to the ellipse at two points on the x-axis. However, the ellipse already intersects the x-axis at those points. So if the circle is tangent to the ellipse at those points, then those points are points of tangency. Therefore, the circle is tangent to the ellipse at (1/√k, 0) and (-1/√k, 0). Therefore, the circle passes through (0, c), (0, -c), (1/√k, 0), and (-1/√k, 0). Wait, but a circle is determined by three non-collinear points. However, here we have four points. Therefore, if the circle passes through all four points, then those four points must lie on the same circle. However, if the circle is tangent at (±1/√k, 0), then those points are points of tangency, so the circle is tangent to the ellipse there, but also passes through the foci. Therefore, maybe the circle passes through the foci and is tangent at those two points on the x-axis. Therefore, the circle passes through (0, c), (0, -c), (1/√k, 0), and (-1/√k, 0). Wait, but if the circle passes through all four points, then those four points must lie on the same circle. Let me check if that's possible.

Alternatively, maybe the circle is tangent at two points on the x-axis, which are not necessarily the same as the ellipse's x-intercepts? Wait, but if the circle is tangent to the ellipse at two points on the x-axis, then those points must lie on both the ellipse and the circle. However, the ellipse intersects the x-axis at (±1/√k, 0). Therefore, if the circle is tangent to the ellipse at those points, then those points are on both the ellipse and the circle. Therefore, the circle must pass through those two points. Therefore, the circle passes through (±1/√k, 0) and (0, ±c). Therefore, the circle passes through four points: (±1/√k, 0) and (0, ±c). Therefore, if these four points lie on the same circle, then we can find the equation of the circle and then use the tangency condition.

Alternatively, maybe the circle is tangent at two points on the x-axis, which are different from the ellipse's x-intercepts? But that seems unlikely because the ellipse only intersects the x-axis at those two points. Therefore, if the circle is tangent to the ellipse at two points on the x-axis, those must be the same points where the ellipse meets the x-axis. Therefore, the circle passes through those two points and the two foci. Therefore, the circle passes through four points: (±1/√k, 0) and (0, ±c). Therefore, we can find the equation of the circle passing through these four points and then use the tangency condition. Wait, but if four points lie on a circle, they must satisfy the circle equation. Let me try to find the equation of the circle passing through these four points.

Alternatively, since the circle is symmetric with respect to both the x-axis and y-axis? Because the points (±1/√k, 0) and (0, ±c) are symmetric with respect to both axes. Therefore, the center of the circle must lie at the origin (0, 0). Wait, is that true? Let me think. If the circle is symmetric with respect to both axes, then yes, the center would be at the origin. However, let me check. Suppose the circle passes through (a, 0), (-a, 0), (0, b), (0, -b). Then, the circle equation would be x² + y² + Dx + Ey + F = 0. But if it passes through (a, 0), then substituting gives a² + 0 + D*a + 0 + F = 0 ⇒ a² + D a + F = 0. Similarly, for (-a, 0): a² - D a + F = 0. Subtracting these two equations: 2 D a = 0 ⇒ D = 0. Similarly, for (0, b): 0 + b² + 0 + E b + F = 0 ⇒ b² + E b + F = 0. For (0, -b): 0 + b² + 0 - E b + F = 0 ⇒ b² - E b + F = 0. Subtracting these two equations: 2 E b = 0 ⇒ E = 0. Therefore, the equation reduces to x² + y² + F = 0. Then, from (a, 0): a² + F = 0 ⇒ F = -a². Therefore, the equation is x² + y² = a². Therefore, the circle is centered at the origin with radius a. Therefore, in our case, the circle passes through (±1/√k, 0) and (0, ±c). Therefore, the radius of the circle would be equal to the distance from the origin to (1/√k, 0), which is 1/√k, and also equal to the distance from the origin to (0, c), which is c. Therefore, we have 1/√k = c. Therefore, c = 1/√k. But earlier, we had c = sqrt(1 - 1/k). Therefore, equating these two expressions:

sqrt(1 - 1/k) = 1/√k.

Let me square both sides:

1 - 1/k = 1/k ⇒ 1 = 2/k ⇒ k = 2.

Wait, so k is 2? Let me check if this is correct. Let me verify.

If k = 2, then the ellipse equation is 2x² + y² = 1. Then, the standard form is x²/(1/2) + y²/1 = 1. Therefore, a² = 1/2, b² = 1. Therefore, since b² > a², the major axis is along the y-axis. Then, c² = b² - a² = 1 - 1/2 = 1/2 ⇒ c = 1/√2. Therefore, the foci are at (0, ±1/√2). The circle passing through these foci and tangent at (±1/√k, 0) = (±1/√2, 0). Therefore, the circle would have radius equal to 1/√2, since the distance from the origin to (1/√2, 0) is 1/√2, and also the distance from the origin to (0, 1/√2) is 1/√2. Therefore, the circle is x² + y² = (1/√2)² = 1/2. Therefore, the equation of the circle is x² + y² = 1/2. Now, check if this circle is tangent to the ellipse at (±1/√2, 0). Let me check the tangency condition.

To check if the circle and the ellipse are tangent at those points, we can check if they have the same tangent line at those points. Let me compute the derivatives.

First, for the ellipse 2x² + y² = 1. Differentiating implicitly:

4x + 2y y' = 0 ⇒ y' = - (4x)/(2y) = -2x/y.

At the point (1/√2, 0), the derivative would be undefined because y = 0. Wait, but that's a problem. Wait, if we plug in y = 0, then the derivative is undefined, which suggests a vertical tangent line. Let me check the ellipse at (1/√2, 0). Since the ellipse is 2x² + y² = 1, at that point, the tangent line is vertical. Similarly, for the circle x² + y² = 1/2, the derivative is:

2x + 2y y' = 0 ⇒ y' = -x/y. At (1/√2, 0), again, the derivative is undefined, suggesting a vertical tangent line. Therefore, both the ellipse and the circle have vertical tangent lines at (±1/√2, 0). Therefore, they are tangent there. Therefore, this seems to satisfy the condition. Therefore, k = 2 is the answer? Wait, but let me check if there are other possibilities. Wait, but according to the previous reasoning, we had that c = 1/√k and c = sqrt(1 - 1/k). Therefore, solving sqrt(1 - 1/k) = 1/√k gives k = 2. Therefore, seems like k = 2 is the answer. However, let me check if there's another way this could happen. For example, maybe the circle is not centered at the origin? Wait, but earlier reasoning suggested that if the circle passes through (±a, 0) and (0, ±c), then it must be centered at the origin. However, is that necessarily true?

Wait, suppose that the circle is not centered at the origin. Let me think. Suppose that the circle passes through (a, 0), (-a, 0), (0, c), (0, -c). Is there a circle not centered at the origin passing through these four points? Let me try to find the equation of such a circle. Let me assume the general equation of a circle: x² + y² + Dx + Ey + F = 0. Then, plugging in (a, 0):

a² + 0 + D*a + 0 + F = 0 ⇒ a² + D a + F = 0. Similarly, (-a, 0): a² - D a + F = 0. Subtracting these equations: 2 D a = 0 ⇒ D = 0. Therefore, D must be zero. Similarly, plugging in (0, c): 0 + c² + 0 + E c + F = 0 ⇒ c² + E c + F = 0. Plugging in (0, -c): 0 + c² + 0 - E c + F = 0 ⇒ c² - E c + F = 0. Subtracting these equations: 2 E c = 0 ⇒ E = 0. Therefore, the equation reduces to x² + y² + F = 0. Then, from (a, 0): a² + F = 0 ⇒ F = -a². Therefore, the equation is x² + y² = a². Therefore, the circle is centered at the origin. Therefore, there is no other circle passing through those four points except the one centered at the origin. Therefore, our previous conclusion that k = 2 is correct. Therefore, the answer is 2. However, let me check again with the tangency condition. Wait, but when I checked the derivatives, both the ellipse and the circle have vertical tangent lines at (±1/√2, 0). Therefore, they are tangent there. Therefore, this seems to satisfy the problem's condition. Therefore, k = 2. However, let me check if there's another possible circle. For example, maybe the circle is tangent at two different points on the x-axis, not necessarily the points where the ellipse meets the x-axis? Wait, but the ellipse only meets the x-axis at those two points. Therefore, if the circle is tangent to the ellipse at two points on the x-axis, those must be the same points. Therefore, the circle must pass through those two points and the foci. Therefore, the previous reasoning holds. Therefore, k = 2. However, let me check with another approach to confirm.

Alternative approach: Let me parametrize the circle. Let me suppose that the circle passes through F₁ = (0, c) and F₂ = (0, -c), and is tangent to the ellipse at points (p, 0) and (-p, 0). Since the ellipse equation is kx² + y² = 1, then at (p, 0), we have kp² + 0 = 1 ⇒ p² = 1/k ⇒ p = ±1/√k. Therefore, the points of tangency are (±1/√k, 0). Therefore, the circle passes through (±1/√k, 0) and (0, ±c). As before, the circle is centered at the origin with radius 1/√k, so equation x² + y² = 1/k. But also, the circle passes through (0, c), so substituting (0, c) into the circle equation: 0 + c² = 1/k ⇒ c² = 1/k. But earlier, we had c² = 1 - 1/k. Therefore, equating:

1 - 1/k = 1/k ⇒ 1 = 2/k ⇒ k = 2. Therefore, same result. Therefore, k = 2. Therefore, seems like the answer is 2. However, let me check if there's a possibility that the circle is not centered at the origin. Wait, but earlier we saw that if the circle passes through (±a, 0) and (0, ±c), then it must be centered at the origin. Therefore, there's no other possibility. Therefore, k = 2 is the answer. Therefore, I think the answer is 2. However, let me check with another method.

Another approach: Let me consider the general equation of a circle passing through F₁ and F₂. Since F₁ and F₂ are (0, c) and (0, -c), the circle passing through these two points must have its center on the x-axis? Wait, no. Wait, the set of circles passing through (0, c) and (0, -c) have their centers lying on the perpendicular bisector of the segment joining (0, c) and (0, -c). The segment joining (0, c) and (0, -c) is vertical, so its perpendicular bisector is the x-axis. Therefore, the center of any circle passing through F₁ and F₂ lies on the x-axis. Therefore, the center is at (h, 0) for some h. Therefore, the circle has equation (x - h)² + y² = r². Now, this circle is also tangent to the ellipse at two points on the x-axis. Let me denote those points as (p, 0) and (-p, 0). Since the circle is tangent to the ellipse at these points, those points lie on both the circle and the ellipse. Also, the tangents at those points are the same for both the circle and the ellipse.

First, let's find the points of tangency. Since they are on the x-axis, their coordinates are (p, 0) and (-p, 0). Since they lie on the ellipse, we have k p² + 0 = 1 ⇒ p² = 1/k ⇒ p = ±1/√k. Therefore, the points are (±1/√k, 0). Therefore, these points lie on the circle. Therefore, substituting (1/√k, 0) into the circle equation:

(1/√k - h)² + 0² = r² ⇒ (1/√k - h)² = r².

Similarly, substituting (-1/√k, 0):

(-1/√k - h)² = r² ⇒ (1/√k + h)² = r².

Therefore, we have:

(1/√k - h)² = (1/√k + h)².

Expanding both sides:

Left side: (1/k - 2h/√k + h²)

Right side: (1/k + 2h/√k + h²)

Setting them equal:

1/k - 2h/√k + h² = 1/k + 2h/√k + h² ⇒ -2h/√k = 2h/√k ⇒ -2h/√k - 2h/√k = 0 ⇒ -4h/√k = 0 ⇒ h = 0.

Therefore, the center of the circle is at (0, 0). Therefore, the circle is centered at the origin. Therefore, the equation is x² + y² = r². Then, since it passes through (1/√k, 0), we have r² = 1/k. Also, since the circle passes through (0, c), we have 0² + c² = r² ⇒ c² = 1/k. But earlier, we had c² = 1 - 1/k. Therefore, equating:

1 - 1/k = 1/k ⇒ 1 = 2/k ⇒ k = 2. Therefore, same result. Therefore, this confirms that k = 2. Therefore, the answer is 2. Therefore, I think this is correct.

But let me check once more with the tangency condition. Suppose k = 2. Then, the ellipse is 2x² + y² = 1. The foci are at (0, ±c) where c² = 1 - 1/2 = 1/2 ⇒ c = 1/√2. The circle is x² + y² = 1/2. Let me check if this circle is tangent to the ellipse at (±1/√2, 0). Let me parametrize the ellipse and the circle. The ellipse at (1/√2, 0) has a vertical tangent line, as we saw earlier. The circle x² + y² = 1/2 also has a vertical tangent line at that point. Therefore, they are tangent there. Therefore, this satisfies the condition. Therefore, k = 2 is correct.

Alternatively, let me check if there's another circle that passes through F₁ and F₂ and is tangent to the ellipse at two points on the x-axis, but not centered at the origin. However, according to the previous reasoning, the center must be at the origin. Therefore, there are no other circles. Therefore, k = 2 is the only solution.

Therefore, I think the answer is 2. Therefore, the value of k is 2.

**Final Answer**
\boxed{2}

To solve the problem, we begin by analyzing the given ellipse:

$$
kx^2 + y^2 = 1
$$

We are told that $ k > 1 $, and the ellipse has foci $ F_1 $ and $ F_2 $, and there exists a circle that passes through these foci and is tangent to the ellipse at two points on the x-axis. Our goal is to find the value of $ k $.

---

### Step 1: Standard Form of the Ellipse

We rewrite the given equation in standard form:

$$
\frac{x^2}{1/k} + \frac{y^2}{1} = 1
$$

This is the standard form of an ellipse centered at the origin. Comparing with the general form:

$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
$$

we identify:

- $ a^2 = \frac{1}{k} $
- $ b^2 = 1 $

Since $ k > 1 $, we have $ a^2 < b^2 $, so the **major axis is along the y-axis**.

---

### Step 2: Coordinates of the Foci

For an ellipse with major axis along the y-axis, the foci are located at:

$$
F_1 = (0, c), \quad F_2 = (0, -c)
$$

where

$$
c^2 = b^2 - a^2 = 1 - \frac{1}{k}
\Rightarrow c = \sqrt{1 - \frac{1}{k}}
$$

---

### Step 3: Points of Tangency on the x-axis

The ellipse intersects the x-axis at points where $ y = 0 $, so:

$$
kx^2 = 1 \Rightarrow x^2 = \frac{1}{k} \Rightarrow x = \pm \frac{1}{\sqrt{k}}
$$

So the ellipse intersects the x-axis at:

$$
\left( \pm \frac{1}{\sqrt{k}}, 0 \right)
$$

The problem states that the circle is **tangent** to the ellipse at these two points. Therefore, the circle must pass through these two points and also through the foci $ F_1 $ and $ F_2 $.

---

### Step 4: Equation of the Circle

Let’s assume the circle is centered at the origin (since the points are symmetric about the origin). Then the general equation of the circle is:

$$
x^2 + y^2 = r^2
$$

This circle passes through:

- $ \left( \frac{1}{\sqrt{k}}, 0 \right) \Rightarrow r^2 = \frac{1}{k} $
- $ (0, c) \Rightarrow r^2 = c^2 = 1 - \frac{1}{k} $

Equating the two expressions for $ r^2 $:

$$
\frac{1}{k} = 1 - \frac{1}{k}
$$

Solving:

$$
\frac{1}{k} + \frac{1}{k} = 1 \Rightarrow \frac{2}{k} = 1 \Rightarrow k = 2
$$

---

### Step 5: Verification

Let’s verify that with $ k = 2 $, the circle is indeed tangent to the ellipse at the x-axis points.

- The ellipse becomes: $ 2x^2 + y^2 = 1 $
- The foci are at $ (0, \pm \sqrt{1 - \frac{1}{2}}) = (0, \pm \frac{1}{\sqrt{2}}) $
- The circle is: $ x^2 + y^2 = \frac{1}{2} $

At the point $ \left( \frac{1}{\sqrt{2}}, 0 \right) $, both the ellipse and the circle pass through this point. The derivative of the ellipse at this point is vertical (since $ y = 0 $), and the derivative of the circle is also vertical (since $ x = \frac{1}{\sqrt{2}} $, $ y = 0 $), so the tangents are the same. Thus, the circle is tangent to the ellipse at these points.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of k for the ellipse given by the equation kx² + y² = 1, where k > 1. The problem states that there's a circle passing through the foci F₁ and F₂ of this ellipse and tangent to the ellipse at two points on the x-axis. Hmm, let me try to break this down step by step.

First, let me recall some properties of ellipses. The standard form of an ellipse is (x²/a²) + (y²/b²) = 1, where a and b are the semi-major and semi-minor axes. However, in this case, the equation is given as kx² + y² = 1. Let me rewrite this in standard form. If I divide both sides by 1, it's already in standard form, but I need to identify a² and b². Let me rearrange:

kx² + y² = 1 ⇒ (x²)/(1/k) + (y²)/1 = 1.

So comparing with the standard form, that would mean that a² = 1/k and b² = 1? Wait, but hold on. Wait, in the standard form, if the major axis is along the x-axis, then a² > b². However, here, since k > 1, then 1/k < 1. Therefore, the major axis is actually along the y-axis? Wait, no. Wait, if the coefficient of x² is k, which is greater than 1, then actually, if we write it as x²/(1/k) + y²/1 = 1, then since 1/k is less than 1, the denominator under x² is smaller, so the major axis is along the y-axis. Wait, but that would mean that the major axis is along the y-axis. However, the problem mentions that the circle is tangent to the ellipse at two points on the x-axis. So maybe the major axis is along the y-axis? Let me confirm.

Alternatively, maybe I should think of the ellipse as (x²)/(1/k) + y²/1 = 1. So if k > 1, then 1/k < 1, so the major axis is along the y-axis. Therefore, the foci are located along the y-axis. Wait, but the problem says that there is a circle passing through F₁ and F₂ and tangent to the ellipse at two points on the x-axis. If the foci are on the y-axis, then the circle passes through two points on the y-axis and is tangent to the ellipse at two points on the x-axis. That seems possible.

But let me check again. Let me recall that for an ellipse, the distance from the center to each focus is c, where c² = a² - b². Wait, but depending on whether the major axis is along x or y. Let me get this straight.

If the standard form is (x²/a²) + (y²/b²) = 1, then if a > b, the major axis is along the x-axis, and c² = a² - b². If b > a, then the major axis is along the y-axis, and c² = b² - a². So in our case, the equation is kx² + y² = 1. Let me write it as x²/(1/k) + y²/1 = 1. Therefore, if we let a² = 1/k and b² = 1, then if k > 1, then 1/k < 1, so b² > a², so the major axis is along the y-axis. Therefore, the foci are located along the y-axis at (0, ±c), where c² = b² - a² = 1 - 1/k. Therefore, c = sqrt(1 - 1/k). Therefore, the foci are at (0, sqrt(1 - 1/k)) and (0, -sqrt(1 - 1/k)). Wait, but the problem says that the circle passes through F₁ and F₂. So the circle passes through these two points on the y-axis. Also, the circle is tangent to the ellipse at two points on the x-axis. So the points of tangency are on the x-axis, meaning their y-coordinates are zero. Let me note that.

So, first, let me note the coordinates of the foci. Let me compute c. Since c² = b² - a² = 1 - 1/k. Therefore, c = sqrt(1 - 1/k). Therefore, the foci are at (0, c) and (0, -c). Therefore, the circle passes through (0, c) and (0, -c). Also, the circle is tangent to the ellipse at two points on the x-axis. Let me denote those points as (p, 0) and (-p, 0), since they are on the x-axis and symmetric with respect to the y-axis. Wait, but the ellipse equation is kx² + y² = 1. If we plug in y = 0, we get kx² = 1 ⇒ x² = 1/k ⇒ x = ±1/√k. Therefore, the ellipse intersects the x-axis at (±1/√k, 0). Therefore, if the circle is tangent to the ellipse at two points on the x-axis, those points must be (±1/√k, 0). Wait, but is that necessarily true? Wait, the circle is tangent to the ellipse at two points on the x-axis. However, the ellipse already intersects the x-axis at those points. So if the circle is tangent to the ellipse at those points, then those points are points of tangency. Therefore, the circle is tangent to the ellipse at (1/√k, 0) and (-1/√k, 0). Therefore, the circle passes through (0, c), (0, -c), (1/√k, 0), and (-1/√k, 0). Wait, but a circle is determined by three non-collinear points. However, here we have four points. Therefore, if the circle passes through all four points, then those four points must lie on the same circle. However, if the circle is tangent at (±1/√k, 0), then those points are points of tangency, so the circle is tangent to the ellipse there, but also passes through the foci. Therefore, maybe the circle passes through the foci and is tangent at those two points on the x-axis. Therefore, the circle passes through (0, c), (0, -c), (1/√k, 0), and (-1/√k, 0). Wait, but if the circle passes through all four points, then those four points must lie on the same circle. Let me check if that's possible.

Alternatively, maybe the circle is tangent at two points on the x-axis, which are not necessarily the same as the ellipse's x-intercepts? Wait, but if the circle is tangent to the ellipse at two points on the x-axis, then those points must lie on both the ellipse and the circle. However, the ellipse intersects the x-axis at (±1/√k, 0). Therefore, if the circle is tangent to the ellipse at those points, then those points are on both the ellipse and the circle. Therefore, the circle must pass through those two points. Therefore, the circle passes through (±1/√k, 0) and (0, ±c). Therefore, the circle passes through four points: (±1/√k, 0) and (0, ±c). Therefore, if these four points lie on the same circle, then we can find the equation of the circle and then use the tangency condition.

Alternatively, maybe the circle is tangent at two points on the x-axis, which are different from the ellipse's x-intercepts? But that seems unlikely because the ellipse only intersects the x-axis at those two points. Therefore, if the circle is tangent to the ellipse at two points on the x-axis, those must be the same points where the ellipse meets the x-axis. Therefore, the circle passes through those two points and the two foci. Therefore, the circle passes through four points: (±1/√k, 0) and (0, ±c). Therefore, we can find the equation of the circle passing through these four points and then use the tangency condition. Wait, but if four points lie on a circle, they must satisfy the circle equation. Let me try to find the equation of the circle passing through these four points.

Alternatively, since the circle is symmetric with respect to both the x-axis and y-axis? Because the points (±1/√k, 0) and (0, ±c) are symmetric with respect to both axes. Therefore, the center of the circle must lie at the origin (0, 0). Wait, is that true? Let me think. If the circle is symmetric with respect to both axes, then yes, the center would be at the origin. However, let me check. Suppose the circle passes through (a, 0), (-a, 0), (0, b), (0, -b). Then, the circle equation would be x² + y² + Dx + Ey + F = 0. But if it passes through (a, 0), then substituting gives a² + 0 + D*a + 0 + F = 0 ⇒ a² + D a + F = 0. Similarly, for (-a, 0): a² - D a + F = 0. Subtracting these two equations: 2 D a = 0 ⇒ D = 0. Similarly, for (0, b): 0 + b² + 0 + E b + F = 0 ⇒ b² + E b + F = 0. For (0, -b): 0 + b² + 0 - E b + F = 0 ⇒ b² - E b + F = 0. Subtracting these two equations: 2 E b = 0 ⇒ E = 0. Therefore, the equation reduces to x² + y² + F = 0. Then, from (a, 0): a² + F = 0 ⇒ F = -a². Therefore, the equation is x² + y² = a². Therefore, the circle is centered at the origin with radius a. Therefore, in our case, the circle passes through (±1/√k, 0) and (0, ±c). Therefore, the radius of the circle would be equal to the distance from the origin to (1/√k, 0), which is 1/√k, and also equal to the distance from the origin to (0, c), which is c. Therefore, we have 1/√k = c. Therefore, c = 1/√k. But earlier, we had c = sqrt(1 - 1/k). Therefore, equating these two expressions:

sqrt(1 - 1/k) = 1/√k.

Let me square both sides:

1 - 1/k = 1/k ⇒ 1 = 2/k ⇒ k = 2.

Wait, so k is 2? Let me check if this is correct. Let me verify.

If k = 2, then the ellipse equation is 2x² + y² = 1. Then, the standard form is x²/(1/2) + y²/1 = 1. Therefore, a² = 1/2, b² = 1. Therefore, since b² > a², the major axis is along the y-axis. Then, c² = b² - a² = 1 - 1/2 = 1/2 ⇒ c = 1/√2. Therefore, the foci are at (0, ±1/√2). The circle passing through these foci and tangent at (±1/√k, 0) = (±1/√2, 0). Therefore, the circle would have radius equal to 1/√2, since the distance from the origin to (1/√2, 0) is 1/√2, and also the distance from the origin to (0, 1/√2) is 1/√2. Therefore, the circle is x² + y² = (1/√2)² = 1/2. Therefore, the equation of the circle is x² + y² = 1/2. Now, check if this circle is tangent to the ellipse at (±1/√2, 0). Let me check the tangency condition.

To check if the circle and the ellipse are tangent at those points, we can check if they have the same tangent line at those points. Let me compute the derivatives.

First, for the ellipse 2x² + y² = 1. Differentiating implicitly:

4x + 2y y' = 0 ⇒ y' = - (4x)/(2y) = -2x/y.

At the point (1/√2, 0), the derivative would be undefined because y = 0. Wait, but that's a problem. Wait, if we plug in y = 0, then the derivative is undefined, which suggests a vertical tangent line. Let me check the ellipse at (1/√2, 0). Since the ellipse is 2x² + y² = 1, at that point, the tangent line is vertical. Similarly, for the circle x² + y² = 1/2, the derivative is:

2x + 2y y' = 0 ⇒ y' = -x/y. At (1/√2, 0), again, the derivative is undefined, suggesting a vertical tangent line. Therefore, both the ellipse and the circle have vertical tangent lines at (±1/√2, 0). Therefore, they are tangent there. Therefore, this seems to satisfy the condition. Therefore, k = 2 is the answer? Wait, but let me check if there are other possibilities. Wait, but according to the previous reasoning, we had that c = 1/√k and c = sqrt(1 - 1/k). Therefore, solving sqrt(1 - 1/k) = 1/√k gives k = 2. Therefore, seems like k = 2 is the answer. However, let me check if there's another way this could happen. For example, maybe the circle is not centered at the origin? Wait, but earlier reasoning suggested that if the circle passes through (±a, 0) and (0, ±c), then it must be centered at the origin. However, is that necessarily true?

Wait, suppose that the circle is not centered at the origin. Let me think. Suppose that the circle passes through (a, 0), (-a, 0), (0, c), (0, -c). Is there a circle not centered at the origin passing through these four points? Let me try to find the equation of such a circle. Let me assume the general equation of a circle: x² + y² + Dx + Ey + F = 0. Then, plugging in (a, 0):

a² + 0 + D*a + 0 + F = 0 ⇒ a² + D a + F = 0. Similarly, (-a, 0): a² - D a + F = 0. Subtracting these equations: 2 D a = 0 ⇒ D = 0. Therefore, D must be zero. Similarly, plugging in (0, c): 0 + c² + 0 + E c + F = 0 ⇒ c² + E c + F = 0. Plugging in (0, -c): 0 + c² + 0 - E c + F = 0 ⇒ c² - E c + F = 0. Subtracting these equations: 2 E c = 0 ⇒ E = 0. Therefore, the equation reduces to x² + y² + F = 0. Then, from (a, 0): a² + F = 0 ⇒ F = -a². Therefore, the equation is x² + y² = a². Therefore, the circle is centered at the origin. Therefore, there is no other circle passing through those four points except the one centered at the origin. Therefore, our previous conclusion that k = 2 is correct. Therefore, the answer is 2. However, let me check again with the tangency condition. Wait, but when I checked the derivatives, both the ellipse and the circle have vertical tangent lines at (±1/√2, 0). Therefore, they are tangent there. Therefore, this seems to satisfy the problem's condition. Therefore, k = 2. However, let me check if there's another possible circle. For example, maybe the circle is tangent at two different points on the x-axis, not necessarily the points where the ellipse meets the x-axis? Wait, but the ellipse only meets the x-axis at those two points. Therefore, if the circle is tangent to the ellipse at two points on the x-axis, those must be the same points. Therefore, the circle must pass through those two points and the foci. Therefore, the previous reasoning holds. Therefore, k = 2. However, let me check with another approach to confirm.

Alternative approach: Let me parametrize the circle. Let me suppose that the circle passes through F₁ = (0, c) and F₂ = (0, -c), and is tangent to the ellipse at points (p, 0) and (-p, 0). Since the ellipse equation is kx² + y² = 1, then at (p, 0), we have kp² + 0 = 1 ⇒ p² = 1/k ⇒ p = ±1/√k. Therefore, the points of tangency are (±1/√k, 0). Therefore, the circle passes through (±1/√k, 0) and (0, ±c). As before, the circle is centered at the origin with radius 1/√k, so equation x² + y² = 1/k. But also, the circle passes through (0, c), so substituting (0, c) into the circle equation: 0 + c² = 1/k ⇒ c² = 1/k. But earlier, we had c² = 1 - 1/k. Therefore, equating:

1 - 1/k = 1/k ⇒ 1 = 2/k ⇒ k = 2. Therefore, same result. Therefore, k = 2. Therefore, seems like the answer is 2. However, let me check if there's a possibility that the circle is not centered at the origin. Wait, but earlier we saw that if the circle passes through (±a, 0) and (0, ±c), then it must be centered at the origin. Therefore, there's no other possibility. Therefore, k = 2 is the answer. Therefore, I think the answer is 2. However, let me check with another method.

Another approach: Let me consider the general equation of a circle passing through F₁ and F₂. Since F₁ and F₂ are (0, c) and (0, -c), the circle passing through these two points must have its center on the x-axis? Wait, no. Wait, the set of circles passing through (0, c) and (0, -c) have their centers lying on the perpendicular bisector of the segment joining (0, c) and (0, -c). The segment joining (0, c) and (0, -c) is vertical, so its perpendicular bisector is the x-axis. Therefore, the center of any circle passing through F₁ and F₂ lies on the x-axis. Therefore, the center is at (h, 0) for some h. Therefore, the circle has equation (x - h)² + y² = r². Now, this circle is also tangent to the ellipse at two points on the x-axis. Let me denote those points as (p, 0) and (-p, 0). Since the circle is tangent to the ellipse at these points, those points lie on both the circle and the ellipse. Also, the tangents at those points are the same for both the circle and the ellipse.

First, let's find the points of tangency. Since they are on the x-axis, their coordinates are (p, 0) and (-p, 0). Since they lie on the ellipse, we have k p² + 0 = 1 ⇒ p² = 1/k ⇒ p = ±1/√k. Therefore, the points are (±1/√k, 0). Therefore, these points lie on the circle. Therefore, substituting (1/√k, 0) into the circle equation:

(1/√k - h)² + 0² = r² ⇒ (1/√k - h)² = r².

Similarly, substituting (-1/√k, 0):

(-1/√k - h)² = r² ⇒ (1/√k + h)² = r².

Therefore, we have:

(1/√k - h)² = (1/√k + h)².

Expanding both sides:

Left side: (1/k - 2h/√k + h²)

Right side: (1/k + 2h/√k + h²)

Setting them equal:

1/k - 2h/√k + h² = 1/k + 2h/√k + h² ⇒ -2h/√k = 2h/√k ⇒ -2h/√k - 2h/√k = 0 ⇒ -4h/√k = 0 ⇒ h = 0.

Therefore, the center of the circle is at (0, 0). Therefore, the circle is centered at the origin. Therefore, the equation is x² + y² = r². Then, since it passes through (1/√k, 0), we have r² = 1/k. Also, since the circle passes through (0, c), we have 0² + c² = r² ⇒ c² = 1/k. But earlier, we had c² = 1 - 1/k. Therefore, equating:

1 - 1/k = 1/k ⇒ 1 = 2/k ⇒ k = 2. Therefore, same result. Therefore, this confirms that k = 2. Therefore, the answer is 2. Therefore, I think this is correct.

But let me check once more with the tangency condition. Suppose k = 2. Then, the ellipse is 2x² + y² = 1. The foci are at (0, ±c) where c² = 1 - 1/2 = 1/2 ⇒ c = 1/√2. The circle is x² + y² = 1/2. Let me check if this circle is tangent to the ellipse at (±1/√2, 0). Let me parametrize the ellipse and the circle. The ellipse at (1/√2, 0) has a vertical tangent line, as we saw earlier. The circle x² + y² = 1/2 also has a vertical tangent line at that point. Therefore, they are tangent there. Therefore, this satisfies the condition. Therefore, k = 2 is correct.

Alternatively, let me check if there's another circle that passes through F₁ and F₂ and is tangent to the ellipse at two points on the x-axis, but not centered at the origin. However, according to the previous reasoning, the center must be at the origin. Therefore, there are no other circles. Therefore, k = 2 is the only solution.

Therefore, I think the answer is 2. Therefore, the value of k is 2.

**Final Answer**
\boxed{2}

To solve the problem, we begin by analyzing the given ellipse:

$$
kx^2 + y^2 = 1
$$

We are told that $ k > 1 $, and the ellipse has foci $ F_1 $ and $ F_2 $, and there exists a circle that passes through these foci and is tangent to the ellipse at two points on the x-axis. Our goal is to find the value of $ k $.

---

### Step 1: Standard Form of the Ellipse

We rewrite the given equation in standard form:

$$
\frac{x^2}{1/k} + \frac{y^2}{1} = 1
$$

This is the standard form of an ellipse centered at the origin. Comparing with the general form:

$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
$$

we identify:

- $ a^2 = \frac{1}{k} $
- $ b^2 = 1 $

Since $ k > 1 $, we have $ a^2 < b^2 $, so the **major axis is along the y-axis**.

---

### Step 2: Coordinates of the Foci

For an ellipse with major axis along the y-axis, the foci are located at:

$$
F_1 = (0, c), \quad F_2 = (0, -c)
$$

where

$$
c^2 = b^2 - a^2 = 1 - \frac{1}{k}
\Rightarrow c = \sqrt{1 - \frac{1}{k}}
$$

---

### Step 3: Points of Tangency on the x-axis

The ellipse intersects the x-axis at points where $ y = 0 $, so:

$$
kx^2 = 1 \Rightarrow x^2 = \frac{1}{k} \Rightarrow x = \pm \frac{1}{\sqrt{k}}
$$

So the ellipse intersects the x-axis at:

$$
\left( \pm \frac{1}{\sqrt{k}}, 0 \right)
$$

The problem states that the circle is **tangent** to the ellipse at these two points. Therefore, the circle must pass through these two points and also through the foci $ F_1 $ and $ F_2 $.

---

### Step 4: Equation of the Circle

Let’s assume the circle is centered at the origin (since the points are symmetric about the origin). Then the general equation of the circle is:

$$
x^2 + y^2 = r^2
$$

This circle passes through:

- $ \left( \frac{1}{\sqrt{k}}, 0 \right) \Rightarrow r^2 = \frac{1}{k} $
- $ (0, c) \Rightarrow r^2 = c^2 = 1 - \frac{1}{k} $

Equating the two expressions for $ r^2 $:

$$
\frac{1}{k} = 1 - \frac{1}{k}
$$

Solving:

$$
\frac{1}{k} + \frac{1}{k} = 1 \Rightarrow \frac{2}{k} = 1 \Rightarrow k = 2
$$

---

### Step 5: Verification

Let’s verify that with $ k = 2 $, the circle is indeed tangent to the ellipse at the x-axis points.

- The ellipse becomes: $ 2x^2 + y^2 = 1 $
- The foci are at $ (0, \pm \sqrt{1 - \frac{1}{2}}) = (0, \pm \frac{1}{\sqrt{2}}) $
- The circle is: $ x^2 + y^2 = \frac{1}{2} $

At the point $ \left( \frac{1}{\sqrt{2}}, 0 \right) $, both the ellipse and the circle pass through this point. The derivative of the ellipse at this point is vertical (since $ y = 0 $), and the derivative of the circle is also vertical (since $ x = \frac{1}{\sqrt{2}} $, $ y = 0 $), so the tangents are the same. Thus, the circle is tangent to the ellipse at these points.

---

### Final Answer

$$
\boxed{2}
$$

Solution:
Writing the equation of the ellipse in the form \[\frac{x^2}{(1/\sqrt k)^2} + \frac{y^2}{1^2} = 1,\]we see that the lengths of the semi-horizontal and semi-vertical axis are $\tfrac{1}{\sqrt{k}}$ and $1,$ respectively. Since $k > 1,$ the vertical axis is the longer (major) axis. Then the distance from the center of the ellipse, the origin, to each focus is \[\sqrt{1 - \left(\sqrt{\frac{1}{k}}\right)^2} = \frac{\sqrt{k-1}}{\sqrt{k}}.\][asy]
size(7cm);
draw((0,-1.4)--(0,1.4),EndArrow); label("$y$",(0,1.4),N);
draw((-1.2,0)--(1.2,0),EndArrow); label("$x$",(1.2,0),E);
draw(xscale(1/sqrt(2))*unitcircle);
draw(scale(1/sqrt(2),1/sqrt(2))*unitcircle);
dot("$F_1$",(0,1/sqrt(2)),NW);
dot("$F_2$",(0,-1/sqrt(2)),SW);
[/asy] The existence of such a circle implies that the origin is equidistant from each focus and each endpoint of the horizontal (minor) axis. Therefore, we have \[\frac{\sqrt{k-1}}{\sqrt{k}} = \frac{1}{\sqrt{k}},\]so $\sqrt{k-1} = 1.$ Thus, $k-1=1,$ and $k=\boxed{2}.$


Explanation:
Okay, so Mr. Brennan has a class with 7 boys and 4 girls. He needs to pick 3 boys and 2 girls for a group presentation. The question is asking how many different ways he can do this. Hmm, let me think.

First, I remember that when we need to choose a certain number of items from a larger set without considering the order, we use combinations. Combinations are different from permutations because with combinations, the order doesn't matter. Since the problem says the order doesn't matter here, combinations are the way to go.

So, the problem is essentially asking for the number of ways to choose 3 boys out of 7 and 2 girls out of 4. Then, since these are independent choices (choosing boys doesn't affect the number of girls available and vice versa), we can multiply the number of ways to choose the boys by the number of ways to choose the girls.

Let me write that down. The number of ways to choose 3 boys from 7 is given by the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items to choose.

So for the boys: C(7, 3). Let me calculate that.

C(7, 3) = 7! / (3! * (7 - 3)!) = 7! / (3! * 4!)

But wait, 7! is 7 × 6 × 5 × 4 × 3 × 2 × 1, and 4! is 4 × 3 × 2 × 1. So maybe I can simplify this before calculating.

7! / 4! is equal to 7 × 6 × 5 × 4! / 4! which cancels out the 4! in the numerator and denominator, leaving 7 × 6 × 5. Then divide by 3! which is 6.

So that would be (7 × 6 × 5) / (3 × 2 × 1) = (210) / 6 = 35.

Okay, so there are 35 ways to choose the boys.

Now for the girls: we need to choose 2 girls out of 4. So that's C(4, 2). Let me compute that.

C(4, 2) = 4! / (2! * (4 - 2)!) = 4! / (2! * 2!)

Again, simplifying. 4! is 4 × 3 × 2 × 1, and 2! is 2 × 1. So 4! / 2! = 4 × 3 × 2! / 2! = 4 × 3 = 12. Then divide by 2! which is 2. So 12 / 2 = 6.

Therefore, there are 6 ways to choose the girls.

Now, since the choices of boys and girls are independent, the total number of ways to choose both is the product of these two numbers. So that's 35 (for the boys) multiplied by 6 (for the girls).

Let me do that multiplication: 35 × 6. Well, 35 × 6 is 210.

Wait, so the total number of ways is 210? Let me just check my calculations again to make sure I didn't make a mistake.

First, for the boys: C(7, 3). Let me verify that.

Another way to compute C(7, 3) is to use the formula step by step.

C(n, k) = n(n - 1)(n - 2)...(n - k + 1) / k!

So for C(7, 3), that would be 7 × 6 × 5 / 3 × 2 × 1. Which is indeed 210 / 6 = 35. That checks out.

For the girls: C(4, 2) is 4 × 3 / 2 × 1 = 12 / 2 = 6. That's correct.

Multiplying 35 × 6: 35 × 6. Let me think. 30 × 6 = 180, and 5 × 6 = 30, so 180 + 30 = 210. Yep, that's right.

So, the total number of ways is 210.

But let me think again if there's another way to approach this problem. Maybe using permutations or something else? Wait, no, because the order doesn't matter. So combinations are definitely the right approach here.

Alternatively, if we thought about arranging the group, but since the order doesn't matter, we don't need to consider permutations.

Another check: suppose we have 7 boys. The number of ways to choose 3 is 35, which seems reasonable. For example, if there were 7 boys, choosing 3, the number is 35. And for 4 girls, choosing 2 is 6. So 35 × 6 is 210.

Alternatively, if I think of the total number of ways to choose 3 boys and 2 girls, it's the same as the product of combinations.

Alternatively, if I thought of the entire class, but no, since we are separating boys and girls.

Wait, another way: suppose we first choose the boys and then the girls. So, for each group of 3 boys, there are 6 ways to choose the girls. So total is 35 × 6.

Alternatively, if I thought of the girls first, then the boys. It would be the same.

Alternatively, if I thought of the total number of ways to choose 5 students, but with the restriction of 3 boys and 2 girls. But that would be similar to the same calculation.

Alternatively, if I thought of the problem as a multinomial coefficient? Wait, but since we are dividing into two groups (boys and girls), but since the groups are distinct (boys and girls), it's just the product of combinations.

Alternatively, if we thought of the problem as first selecting all 5 students, but then considering the gender. But that would be more complicated.

But in this case, since we have a specific number of boys and girls, the straightforward way is to compute combinations for each gender and multiply.

So, I think my answer is correct.

But just to make sure, let me think of a smaller example. Suppose there are 2 boys and 2 girls, and we want to choose 1 boy and 1 girl. How many ways?

Using the same method: C(2,1) * C(2,1) = 2 * 2 = 4. Which is correct: (B1, G1), (B1, G2), (B2, G1), (B2, G2). So that works.

Another example: 3 boys and 2 girls, choose 2 boys and 1 girl.

C(3,2) * C(2,1) = 3 * 2 = 6. Let's check:

Boys: B1, B2, B3. Girls: G1, G2.

Possible groups:

(B1, B2, G1), (B1, B2, G2), (B1, B3, G1), (B1, B3, G2), (B2, B3, G1), (B2, B3, G2). That's 6, which matches.

So the method works.

Therefore, applying the same logic to the original problem, 35 * 6 = 210 is correct.

Therefore, the answer should be 210.

**Final Answer**
The number of ways Mr. Brennan can pick 3 boys and 2 girls is \boxed{210}.

To determine how many ways Mr. Brennan can select 3 boys and 2 girls from his class of 7 boys and 4 girls for a group presentation, we use the concept of **combinations**. Combinations are used when the order of selection does not matter, which is the case here.

---

### Step 1: Choose 3 boys from 7

We use the combination formula:

$$
\binom{n}{k} = \frac{n!}{k!(n-k)!}
$$

So, the number of ways to choose 3 boys from 7 is:

$$
\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
$$

---

### Step 2: Choose 2 girls from 4

Similarly, the number of ways to choose 2 girls from 4 is:

$$
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
$$

---

### Step 3: Multiply the two results

Since the choices of boys and girls are independent, the total number of ways to choose 3 boys and 2 girls is the product of the two combinations:

$$
35 \times 6 = 210
$$

---

### Final Answer

$$
\boxed{210}
$$

[Assistant]
Explanation:
Okay, so Mr. Brennan has a class with 7 boys and 4 girls. He needs to pick 3 boys and 2 girls for a group presentation. The question is asking how many different ways he can do this. Hmm, let me think.

First, I remember that when we need to choose a certain number of items from a larger set without considering the order, we use combinations. Combinations are different from permutations because with combinations, the order doesn't matter. Since the problem says the order doesn't matter here, combinations are the way to go.

So, the problem is essentially asking for the number of ways to choose 3 boys out of 7 and 2 girls out of 4. Then, since these are independent choices (choosing boys doesn't affect the number of girls available and vice versa), we can multiply the number of ways to choose the boys by the number of ways to choose the girls.

Let me write that down. The number of ways to choose 3 boys from 7 is given by the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items to choose.

So for the boys: C(7, 3). Let me calculate that.

C(7, 3) = 7! / (3! * (7 - 3)!) = 7! / (3! * 4!)

But wait, 7! is 7 × 6 × 5 × 4 × 3 × 2 × 1, and 4! is 4 × 3 × 2 × 1. So maybe I can simplify this before calculating.

7! / 4! is equal to 7 × 6 × 5 × 4! / 4! which cancels out the 4! in the numerator and denominator, leaving 7 × 6 × 5. Then divide by 3! which is 6.

So that would be (7 × 6 × 5) / (3 × 2 × 1) = (210) / 6 = 35.

Okay, so there are 35 ways to choose the boys.

Now for the girls: we need to choose 2 girls out of 4. So that's C(4, 2). Let me compute that.

C(4, 2) = 4! / (2! * (4 - 2)!) = 4! / (2! * 2!)

Again, simplifying. 4! is 4 × 3 × 2 × 1, and 2! is 2 × 1. So 4! / 2! = 4 × 3 × 2! / 2! = 4 × 3 = 12. Then divide by 2! which is 2. So 12 / 2 = 6.

Therefore, there are 6 ways to choose the girls.

Now, since the choices of boys and girls are independent, the total number of ways to choose both is the product of these two numbers. So that's 35 (for the boys) multiplied by 6 (for the girls).

Let me do that multiplication: 35 × 6. Well, 35 × 6 is 210.

Wait, so the total number of ways is 210? Let me just check my calculations again to make sure I didn't make a mistake.

First, for the boys: C(7, 3). Let me verify that.

Another way to compute C(7, 3) is to use the formula step by step.

C(n, k) = n(n - 1)(n - 2)...(n - k + 1) / k!

So for C(7, 3), that would be 7 × 6 × 5 / 3 × 2 × 1. Which is indeed 210 / 6 = 35. That checks out.

For the girls: C(4, 2) is 4 × 3 / 2 × 1 = 12 / 2 = 6. That's correct.

Multiplying 35 × 6: 35 × 6. Let me think. 30 × 6 = 180, and 5 × 6 = 30, so 180 + 30 = 210. Yep, that's right.

So, the total number of ways is 210.

But let me think again if there's another way to approach this problem. Maybe using permutations or something else? Wait, no, because the order doesn't matter. So combinations are definitely the right approach here.

Alternatively, if we thought about arranging the group, but since the order doesn't matter, we don't need to consider permutations.

Another check: suppose we have 7 boys. The number of ways to choose 3 is 35, which seems reasonable. For example, if there were 7 boys, choosing 3, the number is 35. And for 4 girls, choosing 2 is 6. So 35 × 6 is 210.

Alternatively, if I think of the total number of ways to choose 3 boys and 2 girls, it's the same as the product of combinations.

Alternatively, if I thought of the entire class, but no, since we are separating boys and girls.

Wait, another way: suppose we first choose the boys and then the girls. So, for each group of 3 boys, there are 6 ways to choose the girls. So total is 35 × 6.

Alternatively, if I thought of the girls first, then the boys. It would be the same.

Alternatively, if I thought of the total number of ways to choose 5 students, but with the restriction of 3 boys and 2 girls. But that would be similar to the same calculation.

Alternatively, if I thought of the problem as a multinomial coefficient? Wait, but since we are dividing into two groups (boys and girls), but since the groups are distinct (boys and girls), it's just the product of combinations.

Alternatively, if we thought of the problem as first selecting all 5 students, but then considering the gender. But that would be more complicated.

But in this case, since we have a specific number of boys and girls, the straightforward way is to compute combinations for each gender and multiply.

So, I think my answer is correct.

But just to make sure, let me think of a smaller example. Suppose there are 2 boys and 2 girls, and we want to choose 1 boy and 1 girl. How many ways?

Using the same method: C(2,1) * C(2,1) = 2 * 2 = 4. Which is correct: (B1, G1), (B1, G2), (B2, G1), (B2, G2). So that works.

Another example: 3 boys and 2 girls, choose 2 boys and 1 girl.

C(3,2) * C(2,1) = 3 * 2 = 6. Let's check:

Boys: B1, B2, B3. Girls: G1, G2.

Possible groups:

(B1, B2, G1), (B1, B2, G2), (B1, B3, G1), (B1, B3, G2), (B2, B3, G1), (B2, B3, G2). That's 6, which matches.

So the method works.

Therefore, applying the same logic to the original problem, 35 * 6 = 210 is correct.

Therefore, the answer should be 210.

**Final Answer**
The number of ways Mr. Brennan can pick 3 boys and 2 girls is \boxed{210}.

To determine how many ways Mr. Brennan can select 3 boys and 2 girls from his class of 7 boys and 4 girls for a group presentation, we use the concept of **combinations**. Combinations are used when the order of selection does not matter, which is the case here.

---

### Step 1: Choose 3 boys from 7

We use the combination formula:

$$
\binom{n}{k} = \frac{n!}{k!(n-k)!}
$$

So, the number of ways to choose 3 boys from 7 is:

$$
\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
$$

---

### Step 2: Choose 2 girls from 4

Similarly, the number of ways to choose 2 girls from 4 is:

$$
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
$$

---

### Step 3: Multiply the two results

Since the choices of boys and girls are independent, the total number of ways to choose 3 boys and 2 girls is the product of the two combinations:

$$
35 \times 6 = 210
$$

---

### Final Answer

$$
\boxed{210}
$$

Solution: There are 4 ways to pick the first girl and 3 ways to pick the second one; however, this counts each pair of girls twice since selecting girl A followed by girl B is the same as selecting girl B followed by girl A, so the total number of ways to pick the girls is $\frac{4\times3}{2}=6$. Similarly, there are 7 ways to pick the first boy, 6 ways to pick the second one, and 5 ways to pick the last one, but this counts each combination of boys 6 times since picking any of the three boys first followed by either of the other two followed by the third one will give the same triplet of boys. So the total number of ways to pick the boys is $\frac{7\times6\times5}{3\times2}=35$, and the total of number of ways to pick the students for the group presentation is $\frac{4\times3}{2}\cdot \frac{7\times6\times5}{3\times2}=\boxed{210}$


Explanation:
Okay, so I need to find the product CD where C and D are integers that satisfy the equation:

$$
\frac{C}{x-3} + \frac{D}{x+8} = \frac{4x - 23}{x^2 + 5x - 24}
$$

for all real values of x except -8 and 3. Alright, let me think about how to approach this.

First, I remember that when you have partial fractions or something like this, you can combine the left side over a common denominator and then equate the numerators. Let me try that.

So, the left side is $\frac{C}{x - 3} + \frac{D}{x + 8}$. To combine these, the common denominator would be (x - 3)(x + 8). Let me write that out:

$$
\frac{C(x + 8) + D(x - 3)}{(x - 3)(x + 8)} = \frac{4x - 23}{x^2 + 5x - 24}
$$

Now, I need to check if the denominator on the right side factors into (x - 3)(x + 8). Let me verify that. Let's multiply (x - 3)(x + 8):

(x - 3)(x + 8) = x^2 + 8x - 3x - 24 = x^2 + 5x - 24.

Yes! That's exactly the denominator on the right side. So, the equation becomes:

$$
\frac{C(x + 8) + D(x - 3)}{(x - 3)(x + 8)} = \frac{4x - 23}{(x - 3)(x + 8)}
$$

Since the denominators are the same, the numerators must be equal for all x (except -8 and 3). Therefore, we can set the numerators equal:

C(x + 8) + D(x - 3) = 4x - 23

Now, let me expand the left side:

C*x + 8C + D*x - 3D = (C + D)x + (8C - 3D)

So, the equation becomes:

(C + D)x + (8C - 3D) = 4x - 23

Since this is true for all x, the coefficients of corresponding powers of x must be equal. Therefore, we can set up a system of equations:

1. Coefficient of x: C + D = 4
2. Constant term: 8C - 3D = -23

So, now I have two equations:

1. C + D = 4
2. 8C - 3D = -23

I need to solve this system for C and D. Let me use substitution or elimination. Maybe elimination is easier here. Let me solve the first equation for D:

D = 4 - C

Then substitute into the second equation:

8C - 3(4 - C) = -23

Let me compute that:

8C - 12 + 3C = -23

Combine like terms:

(8C + 3C) - 12 = -23

11C - 12 = -23

Add 12 to both sides:

11C = -23 + 12

11C = -11

Divide both sides by 11:

C = -1

Now, substitute back into D = 4 - C:

D = 4 - (-1) = 4 + 1 = 5

So, C is -1 and D is 5. Therefore, the product CD is (-1)(5) = -5.

Wait, let me check if that's correct. Let me verify by plugging back into the original equation.

Left side: C/(x - 3) + D/(x + 8) = (-1)/(x - 3) + 5/(x + 8)

Let me combine these:

[ -1*(x + 8) + 5*(x - 3) ] / [(x - 3)(x + 8)]

Compute numerator:

- (x + 8) + 5(x - 3) = -x - 8 + 5x - 15 = ( -x + 5x ) + ( -8 -15 ) = 4x - 23

Which matches the numerator on the right side. So yes, that works. Therefore, C = -1 and D = 5, so CD = -5.

But let me just make sure I didn't make any arithmetic errors. Let me check the solving of the equations again.

Original equations:

1. C + D = 4
2. 8C - 3D = -23

We solved for D = 4 - C, substituted into second equation:

8C - 3*(4 - C) = 8C - 12 + 3C = 11C - 12 = -23

So 11C = -23 + 12 = -11 => C = -1. Then D = 4 - (-1) = 5. Correct.

So CD = (-1)(5) = -5.

Therefore, the answer should be \boxed{-5}.

**Final Answer**
\boxed{-5}

To solve the equation:

$$
\frac{C}{x-3} + \frac{D}{x+8} = \frac{4x - 23}{x^2 + 5x - 24}
$$

for all real values of $ x $ except $ x = -8 $ and $ x = 3 $, we aim to find the product $ CD $, where $ C $ and $ D $ are integers.

---

### Step 1: Factor the Denominator on the Right

We first factor the denominator on the right-hand side:

$$
x^2 + 5x - 24 = (x - 3)(x + 8)
$$

So the equation becomes:

$$
\frac{C}{x - 3} + \frac{D}{x + 8} = \frac{4x - 23}{(x - 3)(x + 8)}
$$

---

### Step 2: Combine the Left-Hand Side

To combine the left-hand side, we find a common denominator:

$$
\frac{C}{x - 3} + \frac{D}{x + 8} = \frac{C(x + 8) + D(x - 3)}{(x - 3)(x + 8)}
$$

Now equating the numerators (since the denominators are the same):

$$
C(x + 8) + D(x - 3) = 4x - 23
$$

---

### Step 3: Expand and Group Like Terms

Expanding the left-hand side:

$$
Cx + 8C + Dx - 3D = (C + D)x + (8C - 3D)
$$

So the equation becomes:

$$
(C + D)x + (8C - 3D) = 4x - 23
$$

Now equating coefficients of $ x $ and the constant terms:

1. $ C + D = 4 $
2. $ 8C - 3D = -23 $

---

### Step 4: Solve the System of Equations

From the first equation:

$$
D = 4 - C
$$

Substitute into the second equation:

$$
8C - 3(4 - C) = -23
$$

$$
8C - 12 + 3C = -23
$$

$$
11C - 12 = -23
$$

$$
11C = -11 \Rightarrow C = -1
$$

Now substitute back to find $ D $:

$$
D = 4 - (-1) = 5
$$

---

### Step 5: Compute the Product $ CD $

$$
CD = (-1)(5) = -5
$$

---

### Final Answer

$$
\boxed{-5}
$$

[Assistant]
Explanation:
Okay, so I need to find the product CD where C and D are integers that satisfy the equation:

$$
\frac{C}{x-3} + \frac{D}{x+8} = \frac{4x - 23}{x^2 + 5x - 24}
$$

for all real values of x except -8 and 3. Alright, let me think about how to approach this.

First, I remember that when you have partial fractions or something like this, you can combine the left side over a common denominator and then equate the numerators. Let me try that.

So, the left side is $\frac{C}{x - 3} + \frac{D}{x + 8}$. To combine these, the common denominator would be (x - 3)(x + 8). Let me write that out:

$$
\frac{C(x + 8) + D(x - 3)}{(x - 3)(x + 8)} = \frac{4x - 23}{x^2 + 5x - 24}
$$

Now, I need to check if the denominator on the right side factors into (x - 3)(x + 8). Let me verify that. Let's multiply (x - 3)(x + 8):

(x - 3)(x + 8) = x^2 + 8x - 3x - 24 = x^2 + 5x - 24.

Yes! That's exactly the denominator on the right side. So, the equation becomes:

$$
\frac{C(x + 8) + D(x - 3)}{(x - 3)(x + 8)} = \frac{4x - 23}{(x - 3)(x + 8)}
$$

Since the denominators are the same, the numerators must be equal for all x (except -8 and 3). Therefore, we can set the numerators equal:

C(x + 8) + D(x - 3) = 4x - 23

Now, let me expand the left side:

C*x + 8C + D*x - 3D = (C + D)x + (8C - 3D)

So, the equation becomes:

(C + D)x + (8C - 3D) = 4x - 23

Since this is true for all x, the coefficients of corresponding powers of x must be equal. Therefore, we can set up a system of equations:

1. Coefficient of x: C + D = 4
2. Constant term: 8C - 3D = -23

So, now I have two equations:

1. C + D = 4
2. 8C - 3D = -23

I need to solve this system for C and D. Let me use substitution or elimination. Maybe elimination is easier here. Let me solve the first equation for D:

D = 4 - C

Then substitute into the second equation:

8C - 3(4 - C) = -23

Let me compute that:

8C - 12 + 3C = -23

Combine like terms:

(8C + 3C) - 12 = -23

11C - 12 = -23

Add 12 to both sides:

11C = -23 + 12

11C = -11

Divide both sides by 11:

C = -1

Now, substitute back into D = 4 - C:

D = 4 - (-1) = 4 + 1 = 5

So, C is -1 and D is 5. Therefore, the product CD is (-1)(5) = -5.

Wait, let me check if that's correct. Let me verify by plugging back into the original equation.

Left side: C/(x - 3) + D/(x + 8) = (-1)/(x - 3) + 5/(x + 8)

Let me combine these:

[ -1*(x + 8) + 5*(x - 3) ] / [(x - 3)(x + 8)]

Compute numerator:

- (x + 8) + 5(x - 3) = -x - 8 + 5x - 15 = ( -x + 5x ) + ( -8 -15 ) = 4x - 23

Which matches the numerator on the right side. So yes, that works. Therefore, C = -1 and D = 5, so CD = -5.

But let me just make sure I didn't make any arithmetic errors. Let me check the solving of the equations again.

Original equations:

1. C + D = 4
2. 8C - 3D = -23

We solved for D = 4 - C, substituted into second equation:

8C - 3*(4 - C) = 8C - 12 + 3C = 11C - 12 = -23

So 11C = -23 + 12 = -11 => C = -1. Then D = 4 - (-1) = 5. Correct.

So CD = (-1)(5) = -5.

Therefore, the answer should be \boxed{-5}.

**Final Answer**
\boxed{-5}

To solve the equation:

$$
\frac{C}{x-3} + \frac{D}{x+8} = \frac{4x - 23}{x^2 + 5x - 24}
$$

for all real values of $ x $ except $ x = -8 $ and $ x = 3 $, we aim to find the product $ CD $, where $ C $ and $ D $ are integers.

---

### Step 1: Factor the Denominator on the Right

We first factor the denominator on the right-hand side:

$$
x^2 + 5x - 24 = (x - 3)(x + 8)
$$

So the equation becomes:

$$
\frac{C}{x - 3} + \frac{D}{x + 8} = \frac{4x - 23}{(x - 3)(x + 8)}
$$

---

### Step 2: Combine the Left-Hand Side

To combine the left-hand side, we find a common denominator:

$$
\frac{C}{x - 3} + \frac{D}{x + 8} = \frac{C(x + 8) + D(x - 3)}{(x - 3)(x + 8)}
$$

Now equating the numerators (since the denominators are the same):

$$
C(x + 8) + D(x - 3) = 4x - 23
$$

---

### Step 3: Expand and Group Like Terms

Expanding the left-hand side:

$$
Cx + 8C + Dx - 3D = (C + D)x + (8C - 3D)
$$

So the equation becomes:

$$
(C + D)x + (8C - 3D) = 4x - 23
$$

Now equating coefficients of $ x $ and the constant terms:

1. $ C + D = 4 $
2. $ 8C - 3D = -23 $

---

### Step 4: Solve the System of Equations

From the first equation:

$$
D = 4 - C
$$

Substitute into the second equation:

$$
8C - 3(4 - C) = -23
$$

$$
8C - 12 + 3C = -23
$$

$$
11C - 12 = -23
$$

$$
11C = -11 \Rightarrow C = -1
$$

Now substitute back to find $ D $:

$$
D = 4 - (-1) = 5
$$

---

### Step 5: Compute the Product $ CD $

$$
CD = (-1)(5) = -5
$$

---

### Final Answer

$$
\boxed{-5}
$$

Solution:
First, we factor the denominator in the right-hand side, to get \[\frac{C}{x - 3} + \frac{D}{x + 8} = \frac{4x - 23}{(x - 3)(x + 8)}.\]We then multiply both sides by $(x - 3)(x + 8)$, to get \[C(x + 8) + D(x - 3) = 4x - 23.\]We can solve for $C$ and $D$ by substituting suitable values of $x$.  For example, setting $x = 3$, we get $11C = -11$, so $C = -1$.  Setting $x = -8$, we get $-11D = -55$, so $D = 5$.  (This may not seem legitimate, because we are told that the given equation holds for all $x$ except $-8$ and $3.$  This tells us that the equation $C(x + 8) + D(x - 3) = 4x - 23$ holds for all $x$, except possibly $-8$ and 3.  However, both sides of this equation are polynomials, and if two polynomials are equal for an infinite number of values of $x$, then the two polynomials are equal for all values of $x$.  Hence, we can substitute any value we wish to into this equation.)

Therefore, $CD = (-1) \cdot 5 = \boxed{-5}$.