Draw a circle that passes point $A$ and is tangent to sides of angle $xoy$ Consider point $A$ in the angle $xoy$ Draw a circle that passes $A$ and is tangent to lines of angle $xoy$.
My attempt:Because it is tangent to both sides of the angle the center of the circle should be on the bisector of the angle.But I can't find a place for the center of the circle that has the same distance from $A$ too.
 A: As noted by you, the center of the circle must lie on the angle bisector of $xoy$. We also note that the center of the circle must be equidistant from the point $A$, and from the line $oy$, and so lies on the parabola with focus $A$ and directrix $oy$.
We thus see that the center of the circle is the intersection of the angle bisector of $xoy$, and the parabola with focus $A$ and directrix $oy$.
The answer to this question shows how to construct the intersection of a line and a parabola with a known focus and directrix.
For concreteness, the process is as follows:
Let $\ell$ be the angle bisector of the angle $xoy$. Reflect $A$ about $\ell$ to obtain the point $A^\prime$. Let $A A^\prime$ intersect $oy$ at $B$. Let $C$ be the intersection of $A A^\prime$ and $\ell$.
Construct semi-circles with diameters $A A^\prime$, and $BC$, and let these intersect at $D$. Construct the circle with center $B$ and radius $BD$. This intersects the line $oy$. Construct a line perpendicular to $oy$ that passes through this point of intersection. This perpendicular line intersects $\ell$ at the center of the required circle.
A: Let $P$, on the angle bisector, be the center of the circle to be drawn. Then the distance of $P$ from one of the angle sides is the same as the distance of $P$ from $A$. It follows that $P$ belongs to the parabola having $A$ as focus and one of the angle sides as directrix. The intersections of that parabola with the angle bisector give then the two only possible positions for $P$.
