Mathematics - Circle Question with Solution | TestHub

Take the point $(3\sec \theta ,2\tan \theta )$. The chord of contact is $3x\sec \theta +2y\tan \theta =9$ (1). If the mid-point is $(h,k)$, the chord is $hx+ky=h^2+k^2$ (2). Comparing (1) and (2): $\sec \theta =\frac{3h}{h^2+k^2}$, $\tan \theta =\frac{9k}{2\left(h^2+k^2\right)}$. Using ${\sec }^2\theta -{\tan }^2\theta =1:\frac{9h^2}{S^2}-\frac{81k^2}{4S^2}=1$ with $S=h^2+k^2$, so $9\left(4h^2-9k^2\right)=4{\left(h^2+k^2\right)}^2$.