Japan Mathematical Olympiad

Given a circle $S$ and a point $Z$ in the plane, we define the power $p=p_S(Z)$ of $Z$ with reference to $S$ in the following way: Let $p=0$ if the point $Z$ lies on the circumference of $S$. Otherwise draw a line through $Z$ and intersecting at the points $I_1$ and $I_2$ with the circle $S$. Then, $p=p_S(Z)=-Z{I}_1\cdot Z{I}_2$ if $Z$ lies in the interior of $S$, and $=Z{I}_1\cdot Z{I}_2$ if $Z$ lies in the exterior of $S$. Here we are considering the line segments $Z{I}_1,Z{I}_2$ to be directed. A well-known theorem (called the theorem on the power of a point) tells us that the value of the power $p$ of $Z$ w.r.t. the circle $S$ is independent of the choice of the line going through it and intersecting the circle $S$, and it is easy to give a proof of this theorem using the fact that angles subtended by an arc of a circle at any pair of points lying on the circle have the same magnitude and similarity of ensuing triangles. Also by considering the line going through $Z$ and the center $O$ of the circle $S$, we see that $p=Z{O}^2-r^2$ holds, where $r$ is the radius of $S$.

Now going back to the problem, we see, from $PA\cdot PC-PB\cdot PD=0$ and the fact that both of the points $X,Y$ lie on both of the circles $S_1,S_2$, that each of the three points $P,X,Y$ satisfies the following: $$p_{S_1}(\text{of the point})+p_{S_2}(\text{of the point})=0(\ast )$$ Let $r_i$ be the radius of the circle $S_i$ and $O_i$ be the center of $S_i$, ($i=1,2$), and denote by $M$ the mid-point of the line segment $O_1{O}_2$. Now for a point $Z$ on the plane note that the following statements are valid: $$\begin{array}{rl}Z\text{ satisfies the condition }(\ast )& ⟺(Z{O}_1^2-r_1^2)+(Z{O}_2^2-r_2^2)=0\\ & ⟺(Z{O}_1^2+Z{O}_2^2)=r_1^2+r_2^2\\ & ⟺2(M{O}_1^2+Z{M}^2)=r_1^2+r_2^2\\ & ⟺Z{M}^2=\frac{1}{2}(r_1^2+r_2^2)-M{O}_1^2.\end{array}$$ Since the right-hand side of the last identity above is independent on $Z$ we conclude that every point in the plane satisfying the condition (*) lies on the circumference of a same circle. In particular, $M$ coincides with the orthocenter of the triangle $PXY$.