Alright, let's tackle this problem step by step. The question is about an ellipse and a circle tangent to it at two points on the x-axis, passing through the foci of the ellipse. We need to find the value of k. Hmm, okay, let's start by recalling some properties of ellipses and circles.

First, the general equation of an ellipse is (x²/a²) + (y²/b²) = 1, where a and b are the semi-major and semi-minor axes. The foci of an ellipse are located at (±c, 0) where c² = a² - b². Since the given equation is kx² + y² = 1, let's rewrite it in the standard form. Dividing both sides by 1 gives kx² + y² = 1, which can be rewritten as (x²)/(1/k) + (y²)/1 = 1. So here, a² = 1/k and b² = 1. Therefore, the semi-major axis is 1 (since b² = 1, so b = 1) and the semi-minor axis is 1/√k. 

Now, the foci of the ellipse are located at (±c, 0), where c² = a² - b². Wait, hold on, actually, in the standard ellipse formula, the major axis is along the x-axis if a > b. But here, since a² = 1/k and b² = 1, we need to check which is larger. Given that k > 1, so 1/k < 1. Therefore, a² < b², which means the major axis is along the y-axis. Wait, that's different. So the major axis is along the y-axis, so the foci are located along the y-axis? Wait, but the problem states that the circle passes through F1 and F2 and is tangent to the ellipse at two points on the x-axis. If the foci are on the y-axis, then the circle passing through both foci would have its center equidistant from both foci, which are on the y-axis. But the circle is tangent to the ellipse at two points on the x-axis. Hmm, that seems tricky. Let me double-check.

Wait, maybe I made a mistake here. Let's go back. The equation is kx² + y² = 1. So comparing to the standard form (x²/a²) + (y²/b²) = 1, we have a² = 1/k and b² = 1. So if a² = 1/k and b² = 1, then the major axis is along the y-axis if b > a, which would be the case here since k > 1. Therefore, the major axis is along the y-axis, and the foci are located at (0, ±c), where c² = b² - a² = 1 - 1/k. So the foci are at (0, ±√(1 - 1/k)). Wait, but the problem mentions F1 and F2, which are on the x-axis? Wait, no, the foci are on the y-axis here. But the circle passes through F1 and F2, which are on the x-axis? Wait, the problem states "F1 and F2 are the foci of the ellipse kx² + y² = 1, where k > 1 is a constant." So the foci are (0, ±√(1 - 1/k)). But then the circle passes through these two foci. If the circle is tangent to the ellipse at two points on the x-axis, which would be (x, 0) points. But the ellipse is kx² + y² = 1. Plugging y = 0, we get x² = 1/k, so the ellipse intersects the x-axis at (±1/√k, 0). So the points of tangency are (1/√k, 0) and (-1/√k, 0). 

So the circle passes through F1 and F2, which are (0, ±√(1 - 1/k)), and is tangent to the ellipse at (±1/√k, 0). Hmm, okay. So the circle passes through the two points on the y-axis (the foci) and is tangent to the ellipse at two points on the x-axis. Let me sketch this mentally. The circle would have to pass through the two foci above and below the x-axis and touch the ellipse at the two x-intercepts. That seems plausible. Let's try to find the equation of such a circle.

A general equation of a circle is x² + y² + Dx + Ey + F = 0. But since the circle passes through (1/√k, 0) and (-1/√k, 0), let's plug those points into the equation. 

First, plug in (1/√k, 0):

(1/√k)² + 0² + D*(1/√k) + E*0 + F = 0

=> 1/k + D/√k + F = 0. Let's call this equation (1).

Similarly, plug in (-1/√k, 0):

(-1/√k)² + 0² + D*(-1/√k) + E*0 + F = 0

=> 1/k - D/√k + F = 0. Let's call this equation (2).

Subtract equation (2) from equation (1):

[1/k + D/√k + F] - [1/k - D/√k + F] = 0 - 0

=> (1/k - 1/k) + (D/√k + D/√k) + (F - F) = 0

=> 2D/√k = 0

=> D = 0.

So the equation of the circle simplifies to x² + y² + Ey + F = 0.

Now, we also know that the circle passes through the two foci of the ellipse, which are (0, ±√(1 - 1/k)). Let's plug in (0, √(1 - 1/k)) into the circle equation:

0² + [√(1 - 1/k)]² + E*√(1 - 1/k) + F = 0

=> (1 - 1/k) + E√(1 - 1/k) + F = 0. Let's call this equation (3).

Similarly, plug in (0, -√(1 - 1/k)):

0² + [ -√(1 - 1/k) ]² + E*(-√(1 - 1/k)) + F = 0

=> (1 - 1/k) - E√(1 - 1/k) + F = 0. Let's call this equation (4).

Now, subtract equation (4) from equation (3):

[ (1 - 1/k) + E√(1 - 1/k) + F ] - [ (1 - 1/k) - E√(1 - 1/k) + F ] = 0 - 0

=> (1 - 1/k - 1 + 1/k) + (E√(1 - 1/k) + E√(1 - 1/k)) + (F - F) = 0

=> 0 + 2E√(1 - 1/k) + 0 = 0

=> 2E√(1 - 1/k) = 0

Since √(1 - 1/k) is not zero (because k > 1), this implies E = 0.

Therefore, the equation of the circle reduces to x² + y² + F = 0. Wait, but that would mean x² + y² = -F, so the radius squared is -F. But radius squared must be positive, so F must be negative. Let's write the equation as x² + y² = -F. Let's call this equation (5).

Now, from equation (1):

1/k + 0 + F = 0 => F = -1/k. So the equation of the circle is x² + y² = 1/k. Wait, that can't be right because the circle passes through (1/√k, 0), which would give (1/√k)^2 + 0 = 1/k, which is 1/k = 1/k. So yes, that works. Similarly, the circle also passes through (-1/√k, 0). So the circle is x² + y² = 1/k. But wait, then the circle's equation is x² + y² = 1/k. Then the radius is 1/√k. But the circle is also supposed to pass through the foci at (0, ±√(1 - 1/k)). Let's check if this is true.

Plug in (0, √(1 - 1/k)) into x² + y² = 1/k:

0 + (√(1 - 1/k))² = 1 - 1/k. So 1 - 1/k must equal 1/k. Therefore:

1 - 1/k = 1/k

=> 1 = 2/k

=> k = 2.

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

If k = 2, then the ellipse equation is 2x² + y² = 1. Then the semi-major axis is 1 (along y-axis), semi-minor axis is 1/√2. The foci are located at (0, ±c) where c² = b² - a² = 1 - 1/2 = 1/2. So c = √(1/2) = 1/√2. Therefore, foci are at (0, ±1/√2). The circle is x² + y² = 1/2. Let's check if (0, ±1/√2) lie on this circle:

x² + y² = 0 + (1/√2)^2 = 1/2, which matches. So the circle passes through the foci. Also, the circle is tangent to the ellipse at (±1/√k, 0) which is (±1/√2, 0). Let's check tangency at (1/√2, 0). The ellipse has derivative dy/dx. Let's compute the derivative of the ellipse 2x² + y² = 1 implicitly:

4x + 2y dy/dx = 0 => dy/dx = -2x/y.

At point (1/√2, 0), the derivative is undefined (since y = 0), which suggests a vertical tangent. The circle x² + y² = 1/2 has derivative dy/dx = -x/y. At (1/√2, 0), the derivative is also undefined, so both curves have vertical tangents there. Therefore, they are tangent at those points. So this seems to check out.

Therefore, k = 2. Wait, but let me make sure there are no other possible solutions. Suppose we didn't make a mistake in assuming E = 0 and D = 0. Let's see.

We had D = 0 from equations (1) and (2), leading to equation of circle as x² + y² + Ey + F = 0. Then plugging in the foci (0, ±√(1 - 1/k)), we found E = 0. Therefore, the circle equation reduces to x² + y² + F = 0, which is x² + y² = -F. Then we used the point (1/√k, 0) to find F = -1/k. Then we checked that the foci lie on the circle, leading to 1 - 1/k = 1/k, so k = 2. 

Alternatively, maybe there's another circle that passes through the foci and is tangent to the ellipse at two points on the x-axis, but not necessarily with the same equation. But according to the steps above, the only solution is k = 2. Let's confirm once again.

Given k = 2, the ellipse is 2x² + y² = 1. The foci are at (0, ±1/√2). The circle is x² + y² = 1/2. This circle passes through the foci and intersects the ellipse at (±1/√2, 0), with vertical tangents there. Therefore, it's tangent. So yes, k = 2 seems correct.

But wait, let me check if there's another possibility. Suppose we didn't assume the circle is centered at the origin. Wait, but the circle passes through (1/√k, 0) and (-1/√k, 0), which are symmetric about the origin, and also passes through the foci (0, ±√(1 - 1/k)), which are also symmetric about the origin. Therefore, the circle must be centered at the origin, because it's symmetric with respect to both x and y axes. Therefore, the circle's equation is x² + y² = r². Then, as before, plugging in (1/√k, 0) gives r² = 1/k, so the circle is x² + y² = 1/k. Then the foci (0, ±√(1 - 1/k)) must lie on this circle, leading to √(1 - 1/k) = ±r. But since r is positive, we have √(1 - 1/k) = 1/√k. Squaring both sides: 1 - 1/k = 1/k => 1 = 2/k => k = 2. So yes, only solution is k = 2. Therefore, the answer is 2.

But let me just verify once more. Suppose k = 2. Then ellipse is 2x² + y² = 1. The foci are at (0, ±1/√2). The circle is x² + y² = 1/2. The points (1/√2, 0) and (-1/√2, 0) lie on the ellipse and the circle. At those points, the ellipse has vertical tangent, as dy/dx is undefined (since y = 0). The circle also has vertical tangent there. Therefore, the circle is tangent to the ellipse at those points. So everything checks out. Therefore, the answer is 2.

**Final Answer**
\boxed{2}
Therefore, the value of $k$ is $\boxed{2}$.
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