Wonderful Circles that leave us confused

Let the midpoint of chord be $(h,k)$ .

Then the equation of chord is $T = S_1$

$\Rightarrow$ $xh +yk -r^2 = h^2 + k^2 -r^2$

$\Rightarrow$ $xh+yk-h^2-k^2$

Since the chord of circle $x^2 +y^2 = r^2$ subtends an angle of $90^{\circ}$ at $(p,q)$ , so the chord must be a diameter of a circle whose center is $(h,k)$ and $(p,q)$ lies on that circle.

Now Equation of family of circle passing through the point of contact between circle $x^2 + y^2 = r^2$ & line $xh +yk -h^2-k^2=0$ is $x^2 + y^2 -r^2 +\lambda (xh +yk -h^2-k^2)=0$ .

Since the center of the circle is $(h,k)$ and from the above equation it is $(-\frac{\lambda h}{2},-\frac{\lambda k}{2})$ ,

$\therefore$ $\lambda = -2$

Now, this circle passes through point $(p,q)$ , so

$p^2 + q^2 -r^2 +\lambda (ph +qk -h^2-k^2)=0$ .

$\Rightarrow$ $p^2 + q^2 -r^2 -2 (ph +qk -h^2-k^2)=0$ .

$\Rightarrow$ $2h^2 +2k^2 +p^2 +q^2 -2ph -2qk-r^2 =0$

Replacing $h$ with $x$ and $k$ with $y$ , we get the locus as $\boxed{2x^2 +2y^2 +p^2 +q^2 -2px -2qy = r^2}$ .

Hence the correct answer is 5.

A One Line Solution (When you solve it it would not take more than 2 lines)

Call Centre of circle O , The point inside the circle As Q , Midpoint Of Chord S And endpoints of the chord as R And T .

We Have Triangle RTQ is a right triangle hence S Would be its circumcentre (Midpoint of hypoteneuse) . So S Would be equidistant From Q,R And T. RS Can Be easily found out using pythagoras theorem in Triangle ROS.Just equate this to the distance between S And Q using distance formula And We Are Done

Shift the coordinate axis to the given point, homogenise (use condition for perpendicularity of pair of straight lines), shift back.

So damn easy!! Thia prob. is for kids