The locus of a point which moves so that the | vedclass

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(D) Let the point be $(x_1, y_1)$.The length of the tangent from $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{x_1^2 + y_1^2 +

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$5x^2 + 5y^2 + 60x + 7 = 0$

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(D) Let the point be $(x_1, y_1)$.The length of the tangent from $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.According to the question,the ratio of the lengths of the tangents is $2:3$:$\frac{\sqrt{x_1^2 + y_1^2 + 4x_1 + 3}}{\sqrt{x_1^2 + y_1^2 - 6x_1 + 5}} = \frac{2}{3}$Squaring both sides,we get:$\frac{x_1^2 + y_1^2 + 4x_1 + 3}{x_1^2 + y_1^2 - 6x_1 + 5} = \frac{4}{9}$Cross-multiplying:$9(x_1^2 + y_1^2 + 4x_1 + 3) = 4(x_1^2 + y_1^2 - 6x_1 + 5)$$9x_1^2 + 9y_1^2 + 36x_1 + 27 = 4x_1^2 + 4y_1^2 - 24x_1 + 20$Rearranging the terms:$5x_1^2 + 5y_1^2 + 60x_1 + 7 = 0$Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $5x^2 + 5y^2 + 60x + 7 = 0$.

The locus of a point which moves so that the ratio of the lengths of the tangents to the circles $x^2 + y^2 + 4x + 3 = 0$ and $x^2 + y^2 - 6x + 5 = 0$ is $2:3$ is

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