Find the locus of the midpoint Find the locus of the midpoint of the chord of the circle $x^2 + y^2=a^2$ which subtends a $90°$ angle at point $(p,q)$ lying inside the circle.
I tried to solve it by taking that let the chord intersect the circle at $(x_1,y_1)$ and $(x_2,y_2)$. Then I found out their slopes and took their product as $-1$. I also tried by taking the lines joining center from chord as perpendicular.
But I couldn't do it. Please tell me a way.
 A: WLOG, let the two extreme points of the chord be $A(a\cos u,a\sin u), B(a\cos v,a\sin v)$
If $P(h,k)$ the midpoint of $A,B$
$2h=a(\cos u+\cos v)\ \ \ \ (1)$
$2k=a(\sin u+\sin v)\ \ \ \ (2)$
$\implies4(h^2+k^2)=a^2\{2+2\cos(u-v)\}\iff\cos(u-v)=?\ \ \ \ (3)$
Now, $AB^2=PA^2+PB^2$
$$\implies4a^2\{(\cos u-\cos v)^2+(\sin u-\sin v)^2\}$$
$$=(p-2a\cos u)^2+(q-2a\sin u)^2+(p-2a\cos v)^2+(q-2a\sin v)^2$$
$\iff4a^2\{2-2\underbrace{\cos(u-v)}\}=2(p^2+q^2)$ 
$-2p\cdot2\underbrace{a(\cos u+\cos v)}-2q\cdot2\underbrace{a(\sin u+\sin v)}$
Use $(1),(2),(3)$ for the under-braced parts.
Can you take it from here?
A: There is another way to solve the problem;
Given the circle equation is... $x^2 + y^2 -a^2$=0;
Let the mid point of the chord be $(h,k)$
Equation of the chord when mid point is given is $ T=S_1$.
$hx+ky-a^2=h^2+k^2-a^2$
or
$$hx+ky-h^2-k^2=0$$
The equation of all cirles passing through the intersection point of the circle and the chord will be
$$x^2+y^2-a^2+\lambda(hx+ky-h^2-k^2)=0$$
Now we have to select a circle from the above equation that should satisfy the given condition.
The circle should subtend right angle by the chord at the point $(p,q)$
This  will be possible when the circle passes through the point and when the chord is diameter of the circle.
 Center of the unique circle should be $(h,k)$ and should pass through the point $(p,q)$.
$$x^2+y^2-a^2+\lambda(hx+ky-h^2-k^2)=0$$
expand 
$x^2 + y^2 +\lambda hx+ \lambda ky....$
This imply the center of the circle is $(-\lambda h/2,-\lambda k/2)$ which should be equal to $(h,k)$.
Thus $\lambda$ is -2.
So,
$$x^2+y^2-a^2+(-2)(hx+ky-h^2-k^2)=0$$
and $x=p$ and $y=q$ since it passes through $(p,q)$
Substitute and simplify
$$p^2 +q^2 -a^2-2hp+-2kq+2h^2+2k^2=0$$
A: In the linked dynamic diagram we see a circle (centered at $O$) and an interior point $P$.  Two perpendicular lines (shown as dashed in the diagram) intersect at $P$, and intercept the circle in four points, which are joined by four chords (shown as boldfaced segments).  So each of the four boldfaced chords subtends a $90°$ angle at $P$.  The midpoints of the four chords are $W,X,Y$ and $Z$, shown in red.
The orientation of the two perpendicular lines can be changed by dragging point $Q$ (purple) around the circumference of the circle.  As $Q$ moves, the chords and their midpoints move as well.  So the question is:
As $Q$ varies, how do the red points move?  What path is traced out by $WX,Y$ and $Z$?
The “Show/Hide Locus” toggle button reveals the path.  It appears to be a circle!  More precisely, it appears to be a circle centered at the midpoint $M$ of the segment joining $O$ to $P$.  You can reveal point $M$ (shown in green) with the second toggle button.
I don’t, unfortunately, have a proof for you of why this is so, but now at least you know what you need to prove.
