Question:
If the line $a x+y=c$, touches both the curves $x^{2}+y^{2}=1$ and $y^{2}=4 \sqrt{2} x$, then $|c|$ is equal to
Correct Option: 4,
Solution:
Equation of tangent on $y^{2}=4 \sqrt{2} x$ is $y t=x+\sqrt{2} t^{2}$
This is also tangent on circle
$\therefore\left|\frac{\sqrt{2} t^{2}}{\sqrt{1+t^{2}}}=1\right| \Rightarrow 2 t^{4}=1+t^{2} \Rightarrow t^{2}=1$
Hence, equation is $\pm y=x+\sqrt{2} \Rightarrow|\mathrm{c}|=\sqrt{2}$