It's definitely a circle, although I'm not sure how you would prove that, or the properties of the circle with respect to the radius of the first circle and the points $ A $ and $ B $.

Let $AX \cap BY = C$. Then by angle formulas, we have that \[\angle{ACB}=\frac{\widehat{AB}-\widehat{XY}}{2}=\frac{\widehat{AB}-90^\circ}{2}\] Since $A$ and $B$ are fixed, $\widehat{AB}$ is constant and hence $\angle{ACB}$ is constant. Hence $C$ lies on the arc of the circle. Geometric continuity can show that all points above the segment $AB$ on this circle are attainable.

Let $P$ be the sought intersection point. Since the configuration is symmetric about the axis of $AB$, we may assume that $P$ lies inside the circle. Note that $\widehat {APB}=\pi-\widehat{YPA}=\pi-\left ( \frac \pi 2-\widehat {AYP}\right ) =\frac \pi 2+\widehat {AYB}$; since $\widehat {AYB}$ is fixed, it follows that the locus of $P$ as $Y$ moves on the circle is the arc of a circle; if $P$ is outside the circle the same reasoning applies (since in that case, letting $P'=AY \cap BX$, $AP'BP$ ic cyclic). We conclude that the desired locus is a circle with chord $AB$; indeed, given a point $P$ on that locus, it suffices to draw the lines $PA$ and $PB$ to find the desired $X$ and $Y$.

Take the center of the circle (0,0) X(-a,0) and Y as (0,a). Now assume the PO of intersection as (h, k) make the equation of the lines through,(-a, 0) (h, k) and (a, 0) (h, k) equating them you will find x=h and y=k put them in x²+y²=r² you will find the circle

Correct me if I'm wrong