Prove the direct common tangent of 2 circles touching externally is the geometric mean of their diameters(meaning that the square of the tangent is the product of the diameters).
Prove that the difference of the squares of the direct and indirect common tangents of 2 non overlapping circles is the product of the 2 diameters.
Thank you in advance.
$$p^2 - q^2 \;=\; \left(\; d^2 - (b-a)^2 \;\right)^2 - \left(\;d^2 - (b+a)^2\;\right)^2 \;=\; 4 a b \;=\; 2a \cdot 2b$$
For tangent circles, one can take $q=0$ in the above to get the geometric mean property. There's also this simple construction showing that the "half-tangent" is the geometric mean of the radii: