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.

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  • $\begingroup$ $K_1$ is not necessarily inside the triangle $ABC$ $\endgroup$
    – Piquito
    Feb 16, 2016 at 13:57

1 Answer 1


This is more of an illustrated and extended comment.

This is a special case of the Problem of Apollonius. You have to expect up to 6 solutions matching your description, e.g. these four:

Illustration of input configuration and four resulting circles

or these six

Illustration of input configuration and six resulting circles

I obtained the above computationally using Lie sphere geometry. But if you read the literature, you'll surely find ruler and compass constructions as well, both for Apollonius' problem in general and for the special case where two circles are in fact lines.

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.


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