If $a 2b 0$, then the positive value of $m$ fro 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

Question

If $a > 2b > 0$, then the positive value of $m$ fro 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

Dr. Abubacker · Accepted Answer

(a)
Given that, any tangent to the circle $x^{2} + y^{2} = b^{2}$ is$y = mx - b\sqrt{1 + m^{2}}$. It touches the circle $(x - a)^{2} + y^{2} = b^{2}$, then
$\frac{ma - b\sqrt{1 + m^{2}}}{\sqrt{m^{2} + 1}} = b$
$\Rightarrow ma = 2b\sqrt{1 + m^{2}}$
$\Rightarrow m^{2}a^{2} = 4b^{2} + 4b^{2}m^{2}$
$	herefore\ \ m = \pm \frac{2b}{\sqrt{a^{2} - 4b^{2}}}$

Q. If $a > 2b > 0$, then the positive value of $m$ fro 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

Explanation

Reviewed by Dr. Abubacker

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