# Given a triangle $ABC$ and a circle $K_1$ tangent to $AB$ and $AC$, construct a circle $K_2$ tangent to $K_1,BA,$ and $BC$

Given a triangle $ABC$ and a circle $K_1$ tangent to $AB$ and $AC$, construct a circle $K_2$ tangent to $K_1,BA,$ and $BC$.

My basic idea is that firstly we know that $K_1$ lies on the angle bisector of $\angle{BAC}$, which is easy to construct. Then in my diagram below we need the perpendicular bisector of $GJ$ to intersect the line through the radical axis of $K_1,K_2$ that passes through both $K_1$ and $K_2$'s centers. The hard part for me is figuring out how to construct this geometrically.

• $K_1$ is not necessarily inside the triangle $ABC$ Feb 16, 2016 at 13:57

The fact that there are always two solutions which touch both $K_1$ and $AB$ in $E$ probably accounts for the fact that you only get 6 solutions, never 8 as would have been possible in the general case. I don't see how this knowledge would make finding these 6 solutions any easier, though.