The product of the slopes of the common tange | vedclass

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(D) Let $S_1: x^2+y^2+2x-2y-2=0$. The center $C_1 = (-1, 1)$ and radius $r_1 = \sqrt{(-1)^2 + 1^2 - (-2)} = \sqrt{4} = 2$.Let $S_2: x^2+y^2-2x+2y+1=0$

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$1$

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(D) Let $S_1: x^2+y^2+2x-2y-2=0$. The center $C_1 = (-1, 1)$ and radius $r_1 = \sqrt{(-1)^2 + 1^2 - (-2)} = \sqrt{4} = 2$.Let $S_2: x^2+y^2-2x+2y+1=0$. The center $C_2 = (1, -1)$ and radius $r_2 = \sqrt{1^2 + (-1)^2 - 1} = \sqrt{1} = 1$.Since the circles are external to each other,the common tangents intersect at a point $P$ which divides the line segment joining the centers $C_1$ and $C_2$ externally in the ratio of their radii $r_1 : r_2 = 2 : 1$.Using the section formula for external division,the coordinates of $P$ are:$P = \left( \frac{2(1) - 1(-1)}{2-1}, \frac{2(-1) - 1(1)}{2-1} \right) = \left( \frac{3}{1}, \frac{-3}{1} \right) = (3, -3)$.Let the equation of the tangent line passing through $P(3, -3)$ be $y + 3 = m(x - 3)$,which simplifies to $mx - y - 3m - 3 = 0$.The perpendicular distance from the center $C_2(1, -1)$ to this line must equal the radius $r_2 = 1$.$\left| \frac{m(1) - (-1) - 3m - 3}{\sqrt{m^2 + (-1)^2}} \right| = 1$ $\Rightarrow \left| \frac{-2m - 2}{\sqrt{m^2 + 1}} \right| = 1$.Squaring both sides: $\frac{4(m+1)^2}{m^2+1} = 1 \Rightarrow 4(m^2 + 2m + 1) = m^2 + 1$.$4m^2 + 8m + 4 = m^2 + 1 \Rightarrow 3m^2 + 8m + 3 = 0$.The product of the slopes $m_1 m_2$ is given by the constant term divided by the coefficient of $m^2$ in the quadratic equation $3m^2 + 8m + 3 = 0$.Therefore,$m_1 m_2 = \frac{3}{3} = 1$.

The product of the slopes of the common tangents drawn to the circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y+1=0$ is:

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