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 $k=\boxed{2}.$
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Let $F_1$ and $F_2$ be the foci of the ellipse $kx^2 + y^2 = 1,$ where $k > 1$ is a constant. Suppose that there is a circle which passes through $F_1$ and $F_2$ and which lies tangent to the ellipse at two points on the $x$-axis. Compute $k.$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
2
Let $F_1$ and $F_2$ be the foci of the ellipse $kx^2 + y^2 = 1,$ where $k > 1$ is a constant. Suppose that there is a circle which passes through $F_1$ and $F_2$ and which lies tangent to the ellipse at two points on the $x$-axis. Compute $k.$
test