IIT-JEE 2012 · Paper 1, Mathematics question 50

Two forms of the same chord. From $(x_1,y_1)$ the chord of contact is $xx_1+yy_1=9$. The chord with midpoint $(h,k)$ is $hx+ky=h^2+k^2$. Comparing,
$$x_1=\frac{9h}{h^2+k^2},\qquad y_1=\frac{9k}{h^2+k^2}.$$

$(x_1,y_1)$ lies on $4x-5y=20$:
$$36h-45k=20(h^2+k^2).$$

Locus. $20(x^2+y^2)-36x+45y=0$.

(A)