If P is a point such that the ratio of the sq | vedclass

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(B) Let the coordinates of $P$ be $(x, y)$.Given the equations of the circles:$C_1: x^2+y^2+2x-4y-20=0$$C_2: x^2+y^2-4x+2y-44=0$The square of the leng

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$(-7,8)$

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(B) Let the coordinates of $P$ be $(x, y)$.Given the equations of the circles:$C_1: x^2+y^2+2x-4y-20=0$$C_2: x^2+y^2-4x+2y-44=0$The square of the length of the tangent from $P(x, y)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $S_1 = x^2+y^2+2gx+2fy+c$.Thus,the squares of the lengths of the tangents are:$T_1^2 = x^2+y^2+2x-4y-20$$T_2^2 = x^2+y^2-4x+2y-44$Given the ratio $\frac{T_1^2}{T_2^2} = \frac{2}{3}$:$\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$This is the equation of a circle in the form $x^2+y^2+2gx+2fy+c=0$,where $2g=14$ and $2f=-16$.Therefore,$g=7$ and $f=-8$.The centre of the circle is $(-g, -f) = (-7, 8)$.

If $P$ is a point such that the ratio of the square of the lengths of the tangents from $P$ to the circles $x^2+y^2+2x-4y-20=0$ 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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