circle

Question: If $a>2b>0$ then the positive value of m for which $y=mx-b\sqrt{1+{{m}^{2}}}$ is a common tangent to ${{x}^{2}}+{{y}^{2}}={{b}^{2}}$ and ${{(x-a)}^{2}}+{{y}^{2}}={{b}^{2}}$ , is



1) $\frac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}$
2) $\frac{\sqrt{{{a}^{2}}-4{{b}^{2}}}}{2b}$
3) $\frac{2b}{a-2b}$
4) $\frac{b}{a-2b}$
Solution: Explanation: No Explanation
Tangent and normal to a circle

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