If a 2b 0,then the positive value of m fo | vedclass

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(A) The equation of any tangent to the circle $x^2 + y^2 = b^2$ is given by $y = mx \pm b\sqrt{1 + m^2}$.Since the given tangent is $y = mx - b\sqrt{1

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$\frac{2b}{\sqrt{a^2 - 4b^2}}$

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(A) The equation of any tangent to the circle $x^2 + y^2 = b^2$ is given by $y = mx \pm b\sqrt{1 + m^2}$.Since the given tangent is $y = mx - b\sqrt{1 + m^2}$,it is a tangent to the first circle.For this line to be a tangent to the second circle $(x - a)^2 + y^2 = b^2$,the perpendicular distance from the center $(a, 0)$ to the line $mx - y - b\sqrt{1 + m^2} = 0$ must be equal to the radius $b$.Thus,$\frac{|ma - 0 - b\sqrt{1 + m^2}|}{\sqrt{m^2 + 1}} = b$.Since $a > 2b$,we have $ma > b\sqrt{1 + m^2}$,so $ma - b\sqrt{1 + m^2} = b\sqrt{m^2 + 1}$.$ma = 2b\sqrt{1 + m^2}$.Squaring both sides,$m^2 a^2 = 4b^2(1 + m^2) = 4b^2 + 4b^2 m^2$.$m^2(a^2 - 4b^2) = 4b^2$.$m^2 = \frac{4b^2}{a^2 - 4b^2}$.Since we need the positive value of $m$,$m = \frac{2b}{\sqrt{a^2 - 4b^2}}$.

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:

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