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 acosθ,asinθ
Then using the midpoint (h,k), the other end of the chord is 2h-acosθ,2k-asinθ
This point lies on the circle x2+y2=a2

 2h-acosθ2+(2k-asinθ)2=a24h2-4ahcosθ+a2cos2θ+4k2-4aksinθ+a2sin2θ=a24h2-4ahcosθ+a2+4k2-4aksinθ=a2 as sin2θ+cos2θ=1ahcosθ+aksinθ=h2+k2   ....1

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

So the line joining (p,q) & acosθ,asinθ and (p,q) & 2h-acosθ,2k-asinθ are perpendicular
So product of slopes = -1

q-2k+asinθp-2h+acosθ×q-asinθp-acosθ=-1q2-2kq-a2sin2θ+2aksinθh2-2hp-a2cos2θ+2ahcosθ=-1q2-2kq-a2sin2θ+2aksinθ = -(h2-2hp-a2cos2θ+2ahcosθ)p2+q2-(2kq+2hp)-a2+2aksinθ+2ahcosθ=0
Using equation (1), we get

So, the locus of the mid-point is

x2+y2-(xp+yq) =12a2-p2-q2

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