The slope of a common tangent to the circles | vedclass

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(B) For the circle $x^2+y^2-4x-8y+16=0$,the center $C_1 = (2, 4)$ and radius $r_1 = \sqrt{2^2+4^2-16} = 2$. For the circle $x^2+y^2-6x-16y+64=0$,the c

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(B) For the circle $x^2+y^2-4x-8y+16=0$,the center $C_1 = (2, 4)$ and radius $r_1 = \sqrt{2^2+4^2-16} = 2$. For the circle $x^2+y^2-6x-16y+64=0$,the center $C_2 = (3, 8)$ and radius $r_2 = \sqrt{3^2+8^2-64} = 3$. Let the common tangent be $mx-y+c=0$. The perpendicular distance from the center to the tangent equals the radius: $\left|\frac{2m-4+c}{\sqrt{1+m^2}}\right| = 2 \Rightarrow c = 2\sqrt{1+m^2}-2m+4$ $\left|\frac{3m-8+c}{\sqrt{1+m^2}}\right| = 3 \Rightarrow c = 3\sqrt{1+m^2}-3m+8$ Equating the two expressions for $c$: $2\sqrt{1+m^2}-2m+4 = 3\sqrt{1+m^2}-3m+8$ $\sqrt{1+m^2} = m-4$ Squaring both sides: $1+m^2 = m^2-8m+16$ $8m = 15 \Rightarrow m = \frac{15}{8}$.

The slope of a common tangent to the circles $x^2+y^2-4x-8y+16=0$ and $x^2+y^2-6x-16y+64=0$ is

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