If the line a x+y=c, touches both the curves


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

  1. (1) 2

  2. (2) $\frac{1}{\sqrt{2}}$

  3. (3) $\frac{1}{2}$

  4. (4) $\sqrt{2}$

Correct Option: 4,


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 $\pm y=x+\sqrt{2} \Rightarrow|\mathrm{c}|=\sqrt{2}$

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