If m_1 and m_2 are the slopes of the direct c | vedclass

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(A) The given circles are $C_1: x^2+y^2-2x-8y+8=0$ with center $C_1(1, 4)$ and radius $r_1 = \sqrt{1^2+4^2-8} = \sqrt{9} = 3$. The second circle is $C

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$-\frac{24}{5}$

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(A) The given circles are $C_1: x^2+y^2-2x-8y+8=0$ with center $C_1(1, 4)$ and radius $r_1 = \sqrt{1^2+4^2-8} = \sqrt{9} = 3$. The second circle is $C_2: x^2+y^2-8x+15=0$ with center $C_2(4, 0)$ and radius $r_2 = \sqrt{4^2+0^2-15} = \sqrt{1} = 1$. The external center of similitude $P$ divides the join of centers $C_1(1, 4)$ and $C_2(4, 0)$ externally in the ratio $r_1:r_2 = 3:1$. $P = \left(\frac{3(4)-1(1)}{3-1}, \frac{3(0)-1(4)}{3-1}\right) = \left(\frac{11}{2}, -2\right)$. Let the slope of the tangent be $m$. The equation of the line passing through $P$ is $y+2 = m(x-\frac{11}{2})$,which simplifies to $2mx - 2y - 11m - 4 = 0$. The distance from $C_2(4, 0)$ to this line is equal to $r_2 = 1$: $\left|\frac{2m(4) - 2(0) - 11m - 4}{\sqrt{(2m)^2 + (-2)^2}}\right| = 1$ $\left|\frac{-3m-4}{\sqrt{4m^2+4}}\right| = 1$ $(3m+4)^2 = 4(m^2+1)$ $9m^2 + 24m + 16 = 4m^2 + 4$ $5m^2 + 24m + 12 = 0$. By the sum of roots formula,$m_1+m_2 = -\frac{24}{5}$.

If $m_1$ and $m_2$ are the slopes of the direct common tangents drawn to the circles $x^2+y^2-2x-8y+8=0$ and $x^2+y^2-8x+15=0$,then $m_1+m_2=$

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