Sum of length of all the common tangents of the circles $x^2+y^2-2x-8y+15=0$ and $x^2+y^2-6x-12y+43=0$ is
Step-by-Step Solution
Key Concept: Circle 1: centre $(1,4)$, $r=\sqrt{2}$. Circle 2: centre $(3,6)$, $r=\sqrt{2}$. Distance $d=2\sqrt{2}=r_1+r_2$: circles are externally tangent.
Sum $=4\sqrt{2}$.
Correct Answer: (C) $4\sqrt{2}$