JEE 2023 — Mathematics PYQ

$$\text{Given parabola is }{y}^2=4x,\text{ where }a=1$$
$$\text{Whose tangent is }y=mx+\frac{a}{m}\Rightarrow y=mx+\frac{1}{m}\dots (i)$$
$$\text{And point of contact is }Q\left(\frac{a}{m^2},\frac{2a}{m}\right)\Rightarrow Q\left(\frac{1}{m^2},\frac{2}{m}\right)$$
$$\text{Since, eqn. (i) is also a tangent of circle }(x-4{)}^2+y^2=16$$
$$\text{So, }\frac{|m\times 4+\frac{1}{m}|}{\sqrt{m^2+1}}=4\Rightarrow \left|\frac{4m^2+1}{m}\right|=4\sqrt{m^2+1}$$
$$\Rightarrow 16m^4+1+8m^2=16m^4+16m^2\Rightarrow 8m^2=1=m=\frac{1}{2\sqrt{2}}$$
$$\text{So }Q(8,4\sqrt{2})\text{ and }PQ=\sqrt{S_1}\Rightarrow (PQ{)}^2=S_1$$
$$=(8-4{)}^2+(4\sqrt{2}{)}^2-16=32$$