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(A) Let the equations of the circles be $S_1 \equiv x^2+y^2-4x-6y-12=0$ and $S_2 \equiv x^2+y^2+6x+18y+26=0$. The length of the tangent from $P(x_1, y

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$5x^2+5y^2-60x-126y-212=0$

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(A) Let the equations of the circles be $S_1 \equiv x^2+y^2-4x-6y-12=0$ and $S_2 \equiv x^2+y^2+6x+18y+26=0$. The length of the tangent from $P(x_1, y_1)$ to a circle $S=0$ is $\sqrt{S}$. Given the ratio of the lengths of the tangents is $2:3$,we have: $\frac{\sqrt{x_1^2+y_1^2-4x_1-6y_1-12}}{\sqrt{x_1^2+y_1^2+6x_1+18y_1+26}} = \frac{2}{3}$ Squaring both sides: $\frac{x_1^2+y_1^2-4x_1-6y_1-12}{x_1^2+y_1^2+6x_1+18y_1+26} = \frac{4}{9}$ Cross-multiplying: $9(x_1^2+y_1^2-4x_1-6y_1-12) = 4(x_1^2+y_1^2+6x_1+18y_1+26)$ $9x_1^2+9y_1^2-36x_1-54y_1-108 = 4x_1^2+4y_1^2+24x_1+72y_1+104$ $5x_1^2+5y_1^2-60x_1-126y_1-212 = 0$ Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $5x^2+5y^2-60x-126y-212=0$.

If $P(x_1, y_1)$ is a point such that the lengths of the tangents from it to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ are in the ratio $2:3$,then the locus of $P$ is

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