<PRE><B>Question 1155987</B>
&lt;br&gt;
Let the triangle be ABC, with right angle at C.&lt;br&gt;
Let P, Q, and R be the points of tangency of the circle with sides AB, BC, and CA, respectively.&lt;br&gt;
We are given that AP=2 and PB=3.&lt;br&gt;
Two tangents from an external point to a circle are congruent, so RC=2 and QB=3.&lt;br&gt;
If r is the radius of the circle, then AC = 2+r and BC = 3+r.&lt;br&gt;
Then in triangle ABC,&lt;br&gt;
{{{(2+r)^2+(3+r)^2 = 5^2}}}&lt;br&gt;
That equation is easily solved to find the radius.&lt;br&gt;
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