JEE Main 2025 — 8 April, Evening Shift — Question 44

Equation of tangent to $y^{2}=g x$ at $(4,6)$ is $3 x-4 y+$ $12=0$

Equation of circle is $(x-4)^{2}+(y-6)^{2}+\lambda(3 x-4 y$ $+12)=0$

$\Rightarrow x^{2}+y^{2}+(3 \lambda-8) x+(-12-4 \lambda) y+52+12 \lambda=0$

$\because \quad 2 \sqrt{g^{2}-c}=0 \Rightarrow g^{2}=c$

$\Rightarrow\left(-\frac{3 \lambda-8}{2}\right)^{2}=52+12 \lambda$

$\Rightarrow 9 \lambda^{2}+64-48 \lambda=208+48 \lambda \Rightarrow 9 \lambda^{2}-96 \lambda-144=0$

$\Rightarrow \lambda=12,-\frac{2}{3} \Rightarrow f=-30,-\frac{14}{3}$

$\Rightarrow r=\sqrt{g^{2}+f^{2}-c}=|f|=|-(2 \lambda+6)|$

$\because$ centre lies in $2^{\text {nd }}$ quadrant

$\Rightarrow 3 \lambda-8>0 \Rightarrow \lambda>\frac{8}{3}$

$\Rightarrow \lambda=12, f=-30, r=30$