We have circles $C_1$ and $C_2$ with centers $(-d,0)$ and $(d,0)$, radii $a_1<d$ and $a_2<d$ respectively. If circle $D$ with radius $r$ (and with centre not necessarily on the x-axis) bisects both $C_1$ and $C_2$, what is the $x$ coordinate of $D$?
My first thoughts on the subject were that if a circle bisects another circle, the line conecting their two centers and one of the points of their intersection will form a right triangle. From here I know that the length of the line connecting the two centers will be $\sqrt {r^2-a_i^2} $ for $i=1,2$.
After this I have been stuck. I have tried using the circle equation but I have been unable to extract useful values for the $x$ coordinate. How can I get the answer?
Edit A curve bisects a circle if it meets its circumference at exactly two points that are diametrically opposite on the circle.
You could use the formula for the equation of a circle given three points:$$\begin{vmatrix}x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \end{vmatrix} = 0.$$
However, to use this formula, you need to know the points of intersection of the circles, which you may not have handy. As you’ve already determined, D’s center is at a distance of $\sqrt(r^2-a_i^2)$ from the center of $C_i$, i.e., it’s at the intersection of the two circles given by the equations $$(x+d)^2+y^2=r^2-a_1^2 \\ (x-d)^2+y^2=r^2-a_2^2.$$ It’s apparent from these two equations that $x$ is independent of $r$, that is, for a given $C_1$ and $C_2$, all circles that bisect the both of them have their centers along a line that’s perpendicular to the line joining the centers of $C_1$ and $C_2$. Eliminating $y$ and $r$ from these equations yields a simple equation for $x$.