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A point P moves such that the lengths of the tangents from it to the circles \mathrm{x^2+y^2=a^2} and \mathrm{x^2+y^2=b^2}are inversely proportional to their radii. The locus of P is a circle of radius

Option: 1

\mathrm{a+b}


Option: 2

\mathrm{\frac{2 a b}{a+b}}


Option: 3

\mathrm{\frac{1}{2} \sqrt{a^2+b^2}}


Option: 4

\mathrm{\sqrt{a^2+b^2}}


Answers (1)

best_answer

Let \mathrm{ P\left(x_1, y_1\right) \Rightarrow \sqrt{S_{11}}=\frac{k}{r}}

\mathrm{ \begin{aligned} & \Rightarrow \sqrt{x_1^2+y_1^2-a^2}=\frac{k}{a} \text { and } \sqrt{x_1^2+y_1^2-b^2}=\frac{k}{b} \\\\ & \therefore a \sqrt{x_1^2+y_1^2-a^2}=b \sqrt{x_1^2+y_1^2-b^2} \\\\ & \Rightarrow a^2\left(x_1^2+y_1^2-a^2\right)=b^2\left(x_1^2+y_1^2-b^2\right) \\\\ & \Rightarrow\left(a^2-b^2\right)\left(x_1^2+y_1^2\right)=a^4-b^4 \\\\ & \Rightarrow x_1^2+y_1^2=a^2+b^2 . \end{aligned} }

The locus of P is a circle \mathrm{x^2+y^2=a^2+b^2\, \, of \, \, radius \, \, \sqrt{a^2+b^2}.}

Posted by

Kuldeep Maurya

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