Find the locus of a point which is such that the length of tangents from it to two concentric circles $x^{2} + y^{2} = a^{2}$ and $x^{2} + y^{2} = b^{2}$ vary inversely with their radii.

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Solution

$(h, k)$ be point so Lenth of tangent to $x^{2} + y^{2} = a^{2}$ is

$\sqrt{h^{2} + k^{2} - a 2}$

and

$x^{2} + y^{2} = b^{2}$ is $\sqrt{h^{2} + k^{2} - b^{2}}$

given

$\frac{\sqrt{h^{2} + k^{2} - b^{2}}}{\sqrt{h^{2} + k^{2} - a^{2}}} = \frac{a}{b}$

square both side

$b^{2} (h^{2} + k^{2} - b^{2}) = a^{2} (h^{2} + k^{2} - a^{2})$

$(b^{2} - a^{2}) (h^{2} + k^{2}) = b^{4} - a^{4}$

thus $h^{2} + k^{2} = a^{2} + b^{2}$

so $x^{2} + y^{2} = a^{2} + b^{2}$ is the locus

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