Question

Prove that the locus of the mid-points of the chords of circle $x^2+y^2=4$ such that the segment intercepted by the chord on the curve $x^2=2(x+y)$ subtends a right angle at the origin is $x^2+y^2-2x-2y=0$

Answer 1

If $(h,k)$ be mid-point, then $T=S_1$
or $hx+ky=h^2+k^2......\left(1\right)$
Now make the curve $x^2=2(x+y)$ homogeneous by the help of $\left(1\right)$
$\therefore x^2=2(x+y)\cdot \frac{hx+ky}{h^2+k^2}$
or $x^2(h^2+k^2)-2(hx+ky)(x+y)=0$
It represents a pair of perpendicular lines through origin $\therefore A+B=0$
or $(h^2+k^2-2h)-2k=0$
Now generalize $(h,k)$.