If the lengths of the tangents drawn from P t | vedclass

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(D) We know that the length of the tangent drawn from $(x_1, y_1)$ to the circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.Let $P =

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$x^2+y^2-2x-12y+8=0$

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(D) We know that the length of the tangent drawn from $(x_1, y_1)$ to the circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.Let $P = (h, k)$.According to the question,the ratio of the lengths of the tangents is $2:1$:$\frac{\sqrt{h^2+k^2-2h+4k-20}}{\sqrt{h^2+k^2-2h-8k+1}} = \frac{2}{1}$Squaring both sides:$\frac{h^2+k^2-2h+4k-20}{h^2+k^2-2h-8k+1} = 4$$h^2+k^2-2h+4k-20 = 4(h^2+k^2-2h-8k+1)$$h^2+k^2-2h+4k-20 = 4h^2+4k^2-8h-32k+4$$3h^2+3k^2-6h-36k+24 = 0$Dividing by $3$:$h^2+k^2-2h-12k+8 = 0$Therefore,the locus of point $P$ is $x^2+y^2-2x-12y+8=0$.

If the lengths of the tangents drawn from $P$ to the circles $x^2+y^2-2x+4y-20=0$ and $x^2+y^2-2x-8y+1=0$ are in the ratio $2:1$,then the locus of $P$ is

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