A) \[1/2\]
B) \[2\]
C) \[\sqrt{2}\]
D) \[\frac{1}{\sqrt{2}}\]
Correct Answer: C
Solution :
Tangent to\[~{{y}^{2}}=4\sqrt{2}x\] is \[y=mx+\frac{\sqrt{2}}{m}\]it is also tangent to \[{{x}^{2}}+{{y}^{2}}=1\] \[\Rightarrow \left| \frac{\sqrt{2}/m}{\sqrt{1+{{m}^{2}}}} \right|=1\Rightarrow m=\pm 1\] \[\Rightarrow \]Tagent will be \[y=x+\sqrt{2}\] or \[y=-x-\sqrt{2}\]compare with \[y=ax+C\] \[\Rightarrow a=\pm 1\,\,\,\And \,\,\,C=\pm \sqrt{2}\]
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