If $P (x_{1}, y_{1})$ is a point such that the lengths of the tangents from it to the circles $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ are in the ratio $2 : 3$ then the locus of $P$ is

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Solution

Length of tangents = $\sqrt{x_{1}^{1} + x_{1}^{2} - 4 x_{1} - 6 y_{1} - 12}$ & $\sqrt{x_{1}^{2} + y_{1}^{2} + 6 x_{1} + 18 y_{1} + 26}$
Given $\Rightarrow \sqrt{\frac{x_{1}^{2} + y_{1}^{2} - 4 x_{1} - 6 y_{1} - 12}{x_{1}^{2} + y_{1}^{2} = 6 x_{1} + 18 y_{1} + 26}} = \frac{2}{3} \Rightarrow \frac{4}{9}$
$\frac{4}{9} = \frac{x_{1}^{2} + y_{1}^{2} - 4 x_{1} - 6 y_{1} - 12}{x_{1}^{2} + y_{1}^{2} + 6 x_{1} + 18 y_{1} + 26} \Rightarrow 9 x_{1}^{2} + 9 y_{1}^{2} - 36 x_{1} - 54 y_{1} - 108 = 4 x_{1}^{2} + 4 y_{1}^{2} + 24 x_{1} + 72 y_{1} + 104$
$5 x_{1}^{2} + 5 y_{1}^{2} - 60 x_{1} - 126 y_{1} - 212 = 0$
$x^{2} + y^{2} - 12 x - \frac{126}{5} y - \frac{212}{5} = 0$

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If P(x1,y1) is a point such that the lengths of the tangents from it to the circles x2+y2−4x−6y−12=0 and x2+y2+6x+18y+26=0 are in the ratio 2:3 then the locus of P is

If $P (x_{1}, y_{1})$ is a point such that the lengths of the tangents from it to the circles $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ are in the ratio $2 : 3$ then the locus of $P$ is