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: , 3
Solution:
Tangent to $\mathrm{y}^{2}=4 \sqrt{2} \mathrm{x}$ is $\mathrm{y}=\mathrm{mx}+\frac{\sqrt{2}}{\mathrm{~m}}$
it is also tangent to $x^{2}+y^{2}=1$
$\Rightarrow\left|\frac{\sqrt{2} / \mathrm{m}}{\sqrt{1+\mathrm{m}^{2}}}\right|=1 \Rightarrow \mathrm{m}=\pm 1$
$\Rightarrow$ Tagent will be $y=x+\sqrt{2}$ or $y=-x-\sqrt{2}$
compare with $y=-a x+C$
$\Rightarrow \mathrm{a}=\pm 1 \& \mathrm{C}=\pm \sqrt{2}$
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