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If $P$ is a point such that the ratio of the square of the lengths of the tangents from $P$ to the circles $x^2+y^2+2 x-4 y-20=0$ and $x^2+y^2-4 x+2 y-2 y-44=0$ is $2: 3$, then the locus of $P$ is a circle with centre :
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Verified Answer
The correct answer is:
$(-7,8)$
Let co-ordinates of $P$ be $\left(x_1, y_1\right)$.
Given that,
$x^2+y^2+2 x-4 y-20=0$ ...(i)
and $\quad x^2+y^2-4 x+2 y-44=0$ ...(ii)
Length of the tangent from $P$ to Eq. (i)
$=x_1^2+y_1^2+2 x_1-4 y_1-20$ ...(iii)
Length of the tangent from $P$ to Eq. (ii)
$=x_1^2+y_1^2-4 x_1+2 y_1-44$ ...(iv)
Given that ratio of lengths of tangent $=\frac{2}{3}$
$\Rightarrow \quad \frac{x_1^2+y_1^2+2 x_1-4 y_1-20}{x_1^2+y_1^2-4 x_1+2 y_1-44}=\frac{2}{3}$
$\Rightarrow \quad 3 x_1^2+3 y_1^2+6 x_1-12 y_1-60$ $=2 x_1^2+2 y_1^2-8 x_1+4 y_1-88$
$\Rightarrow \quad x_1^2+y_1^2+14 x_1-16 y_1+28=0$
$\therefore$ Locus of points $P$ is
$x^2+y^2+14 x-16 y+28=0$
Centre of the circle is $(-7,8)$.
Given that,
$x^2+y^2+2 x-4 y-20=0$ ...(i)
and $\quad x^2+y^2-4 x+2 y-44=0$ ...(ii)
Length of the tangent from $P$ to Eq. (i)
$=x_1^2+y_1^2+2 x_1-4 y_1-20$ ...(iii)
Length of the tangent from $P$ to Eq. (ii)
$=x_1^2+y_1^2-4 x_1+2 y_1-44$ ...(iv)
Given that ratio of lengths of tangent $=\frac{2}{3}$
$\Rightarrow \quad \frac{x_1^2+y_1^2+2 x_1-4 y_1-20}{x_1^2+y_1^2-4 x_1+2 y_1-44}=\frac{2}{3}$
$\Rightarrow \quad 3 x_1^2+3 y_1^2+6 x_1-12 y_1-60$ $=2 x_1^2+2 y_1^2-8 x_1+4 y_1-88$
$\Rightarrow \quad x_1^2+y_1^2+14 x_1-16 y_1+28=0$
$\therefore$ Locus of points $P$ is
$x^2+y^2+14 x-16 y+28=0$
Centre of the circle is $(-7,8)$.
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