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


A) \( \Large \frac{2b}{\sqrt{a^{2}-4b^{2}}} \)

B) \( \Large \frac{\sqrt{a^{2}-4b^{2}}}{2b} \)

C) \( \Large \frac{2b}{a-2b} \)

D) \( \Large \frac{b}{a-2b} \)

Correct Answer:
A) \( \Large \frac{2b}{\sqrt{a^{2}-4b^{2}}} \)

Description for Correct answer:
Given that, any tangent to the circle \( \Large x^{2}+y^{2}=b^{2}\ is\ y=mx-b\sqrt{1+m^{2}} \) It touches the circle \( \Large \left(x-a\right)^{2}+y^{2}=b^{2} \) ,

then \( \Large \frac{ma-b\sqrt{1+m^{2}}}{\sqrt{m^{2}+1}} = b \)

=> \( \Large ma = 2b\sqrt{1+m^{2}} \)

=> \( \Large m^{2}a^{2} = 4b^{2}+4b^{2}m^{2} \)

\( \Large \therefore\ m = \pm \frac{2b}{\sqrt{a^{2}-4b^{2}}} \)

Part of solved Circles questions and answers : >> Elementary Mathematics >> Circles




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