# 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

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

Using equation (1), we get

$\left({\mathrm{p}}^{2}+{\mathrm{q}}^{2}\right)-2\left(\mathrm{kq}+\mathrm{hp}\right)-{\mathrm{a}}^{2}+2{\mathrm{h}}^{2}+2{\mathrm{k}}^{2}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2\left({\mathrm{h}}^{2}+{\mathrm{k}}^{2}\right)-2\left(\mathrm{kq}+\mathrm{hp}\right)={\mathrm{a}}^{2}-\left({\mathrm{p}}^{2}+{\mathrm{q}}^{2}\right)\phantom{\rule{0ex}{0ex}}$
So, the locus of the mid-point is

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