If P(x, y) is a point such that the ratio of | vedclass

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(B) Let the point $P$ be $(x, y)$. The square of the length of the tangent from $P(x, y)$ to a circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ is given by

Vedclass · Accepted Answer

$(-7, 8)$

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(B) Let the point $P$ be $(x, y)$. The square of the length of the tangent from $P(x, y)$ to a circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $S_1 = x^2 + y^2 + 2gx + 2fy + c$.Given the ratio of the squares of the lengths of the tangents is $2 : 3$:$\frac{x^2 + y^2 + 2x - 4y - 20}{x^2 + y^2 - 4x + 2y - 44} = \frac{2}{3}$$3(x^2 + y^2 + 2x - 4y - 20) = 2(x^2 + y^2 - 4x + 2y - 44)$$3x^2 + 3y^2 + 6x - 12y - 60 = 2x^2 + 2y^2 - 8x + 4y - 88$$x^2 + y^2 + 14x - 16y + 28 = 0$Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = 14 \Rightarrow g = 7$ and $2f = -16 \Rightarrow f = -8$.The centre of the circle is $(-g, -f) = (-7, 8)$.

If $P(x, y)$ is a point such that the ratio of the squares of the lengths of the tangents from $P$ to the circles $x^2 + y^2 + 2x - 4y - 20 = 0$ and $x^2 + y^2 - 4x + 2y - 44 = 0$ is $2 : 3$,then the locus of $P$ is a circle with centre

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