Find the locus of mid point of chord of the circle x^2+y^2=a^2 which subtends a 90 degree angle at point (p,q) lying inside the circle.

Dear Student

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

Let the one end of the chord be $\left(\mathrm{acos\theta },\mathrm{asin\theta }\right)$
Then using the midpoint (h,k), the other end of the chord is $\left(2\mathrm{h}-\mathrm{acos\theta },2\mathrm{k}-\mathrm{asin\theta }\right)$
This point lies on the circle $x^2+y^2=a^2$
So

 $$\begin{gathered} {\left(2\mathrm{h}-\mathrm{acos\theta }\right)}^2+(2\mathrm{k}-\mathrm{asin\theta }{)}^2={\mathrm{a}}^2 \\ 4{\mathrm{h}}^2-4\mathrm{ahcos\theta }+{\mathrm{a}}^2{\cos }^2\mathrm{\theta }+4{\mathrm{k}}^2-4\mathrm{aksin\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }={\mathrm{a}}^2 \\ 4{\mathrm{h}}^2-4\mathrm{ahcos\theta }+{\mathrm{a}}^2+4{\mathrm{k}}^2-4\mathrm{aksin\theta }={\mathrm{a}}^2 \mathrm{as} {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1 \\ \mathrm{ahcos\theta }+\mathrm{aksin\theta }={\mathrm{h}}^2+{\mathrm{k}}^{2 }....\left(1\right) \end{gathered}$$

Also the chord subtends right angle at (p,q).

So the line joining (p,q) & $\left(\mathrm{acos\theta },\mathrm{asin\theta }\right)$ and (p,q) & $\left(2\mathrm{h}-\mathrm{acos\theta },2\mathrm{k}-\mathrm{asin\theta }\right)$ are perpendicular
So product of slopes = -1

$$\begin{gathered} \frac{\left(\mathrm{q}-2\mathrm{k}+\mathrm{asin\theta }\right)}{\left(\mathrm{p}-2\mathrm{h}+\mathrm{acos\theta }\right)}\times \frac{\left(\mathrm{q}-\mathrm{asin\theta }\right)}{\left(\mathrm{p}-\mathrm{acos\theta }\right)}=-1 \\ \frac{{\mathrm{q}}^2-2\mathrm{kq}-{\mathrm{a}}^2{\sin }^2\mathrm{\theta }+2\mathrm{aksin\theta }}{{\mathrm{h}}^2-2\mathrm{hp}-{\mathrm{a}}^2{\cos }^2\mathrm{\theta }+2\mathrm{ahcos\theta }}=-1 \\ {\mathrm{q}}^2-2\mathrm{kq}-{\mathrm{a}}^2{\sin }^2\mathrm{\theta }+2\mathrm{aksin\theta } = -({\mathrm{h}}^2-2\mathrm{hp}-{\mathrm{a}}^2{\cos }^2\mathrm{\theta }+2\mathrm{ahcos\theta }) \\ \left({\mathrm{p}}^2+{\mathrm{q}}^2\right)-(2\mathrm{kq}+2\mathrm{hp})-{\mathrm{a}}^2+2\mathrm{aksin\theta }+2\mathrm{ahcos\theta }=0 \end{gathered}$$
Using equation (1), we get

$$\begin{gathered} ({\mathrm{p}}^2+{\mathrm{q}}^2)-2(\mathrm{kq}+\mathrm{hp})-{\mathrm{a}}^2+2{\mathrm{h}}^2+2{\mathrm{k}}^2=0 \\ 2({\mathrm{h}}^2+{\mathrm{k}}^2)-2(\mathrm{kq}+\mathrm{hp})={\mathrm{a}}^2-({\mathrm{p}}^2+{\mathrm{q}}^2) \end{gathered}$$
So, the locus of the mid-point is

${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{(}\mathbf{xp}\mathbf{+}\mathbf{yq}\mathbf{)}\mathbf{ }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left({\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{p}}^{\mathbf{2}}\mathbf{-}{\mathbf{q}}^{\mathbf{2}}\right)$