MathLabs

Problem 5

Two tangent circles and a point P on their common tangent perpendicular to the line joining their centres are given. Construct with ruler and compass all circles tangent to the two given circles and passing through P.
Step 3 of 5: Transform a sought circle
C∋P→IPt,t common tangent to C1,C2.C\ni P\xrightarrow{\mathcal I_P}t,\qquad t\text{ common tangent to }C_1,C_2.
Detailed analysis

A circle through the inversion centre becomes a line. Tangency is preserved, and the invariant images of the two given circles remain the same, so the image line is a common tangent of the two given circles. It cannot be PT because PT is already fixed and is the tangent containing P in the original configuration.