If P(x_1, y_1) is a point such that the lengt | vedclass

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(B) Let the point $P$ be $(x, y)$. The length of the tangent from a point $(x, y)$ to a circle $S=0$ is $\sqrt{S}$.Given circles are $S_1 \equiv x^2+y

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$x^2+y^2-12x-\frac{126}{5}y-\frac{212}{5}=0$

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(B) Let the point $P$ be $(x, y)$. The length of the tangent from a point $(x, y)$ to a circle $S=0$ is $\sqrt{S}$.Given circles are $S_1 \equiv x^2+y^2-4x-6y-12=0$ and $S_2 \equiv x^2+y^2+6x+18y+26=0$.The ratio of the lengths of the tangents is $\frac{\sqrt{S_1}}{\sqrt{S_2}} = \frac{2}{3}$.Squaring both sides,we get $\frac{S_1}{S_2} = \frac{4}{9}$,which implies $9S_1 = 4S_2$.Substituting the equations of the circles:$9(x^2+y^2-4x-6y-12) = 4(x^2+y^2+6x+18y+26)$.$9x^2+9y^2-36x-54y-108 = 4x^2+4y^2+24x+72y+104$.$5x^2+5y^2-60x-126y-212 = 0$.Dividing by $5$,we get the locus of $P$ as $x^2+y^2-12x-\frac{126}{5}y-\frac{212}{5}=0$.

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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