# Equation of a circle with following conditions

Say I have a circle with radius $r$ and center $(c_1,c_2)$. What values for those constants makes the circle intersect the points $(-a,0)$ and $(a,0)$ and the upper plane part of the circle has length $l$? My attempt begins by setting $c_1 = 0$ since I figure the circle must be symmetric about the y-axis. However, after this I am having trouble determining $r$ and $c_2$.

To start, the equation of the circle (writing $(c, d)$ instead of $(c_1, c_2)$),is $(x-c)^2+(y-d)^2 = r^2$.
Since it goes through $(a, 0)$ and $(-a, 0)$, $r^2 = (a-c)^2+d^2 = (-a-c)^2+d^2$. So $a^2-2ac+c^2+d^2 =a^2+2ac+c^2+d^2$ or, subtracting, $4ac=0$. If $a \ne 0$, then $c = 0$, so the circle is $x^2+(y-d)^2 = r^2$.
Also, $r^2 = a^2+d^2$ so the circle is $x^2+(y-d)^2 = a^2+d^2$.
The part in the upper plane is the part above $y = 0$. Since the circle passes through $(-a, 0)$ and $(a, 0)$, the part of the circle in the upper plane passes through these points and, setting $x=0$, $(y-d)^2 =a^2+d^2$ or $y = d+\sqrt{a^2+d^2}$ (we use the positive root since we want $y > 0$ for the upper plane).
The final step would be to get an expression for the length of this arc of the circle (in terms of $a$ and $d$), set this expression equal to $l$, and solve for $d$ in terms of $a$ and $l$.