JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Circle

Let $P(h,k)$ be the midpoint of the chord of contact. The equation of the chord bisected at $P$ is $hx+ky=h^2+k^2$. For a point $(\alpha ,\frac{4\alpha -20}{5})$ on the line, the chord of contact is $\alpha x+(\frac{4\alpha -20}{5})y=9$. Comparing coefficients: $\frac{h}{\alpha }=\frac{k}{\frac{4\alpha -20}{5}}=\frac{h^2+k^2}{9}$. From the first two ratios, $\alpha =\frac{20h}{4h-5k}$. Substituting into the third ratio: $\frac{h(4h-5k)}{20h}=\frac{h^2+k^2}{9}$, which simplifies to $20(h^2+k^2)=9(4h-5k)$, or $20(x^2+y^2)-36x+45y=0$.