A circle $Γ_1$ of radius $25$ is externally tangent to a circle $Γ_2$ of radius $16$ at $C$. Let $AB$ be a common direct tangent, so that $A$ lies on $Γ_1$ and $B$ lies on $Γ_2$. Draw the tangent to $Γ_1$ that is parallel to $AB$; let this tangent intersect $Γ_1$ at $T$, and the common transverse tangent through C at $U$. Then find the length of $TU $. From where do I start the problem. I tried drawing some perpendiculars, but I don't think it helps.
On
Direct trig calculations after a sketch is drawn. The circle radii are $ (R,r)= (25,16)$.
The half -angle between direct tangents $= \theta~ $ is given by
$$ \sin \theta =\frac{R+r}{R-r}\to \theta=12.680383^{\circ}$$
$\angle FHA $ is half the complement of $\theta$
$$ TU= FH= R \cot (45^{\circ}- \theta/2)= 25/ \tan 38.66^{\circ}\approx 31.25$$
which tallies with Geogebra construction:
Assume that $CU$ and $AB$ meet at $V\in AB$. And $O_i$ is a center of $\Gamma_i$.
(1) Note that $$O_1TUC\ {\rm is\ similar\ to}\ O_2BVC$$ And note that the ratio is $16/25$. That is if we know $BV=CV=x$, then we are done.
(2) In $O_1O_2BA$, $$\angle AO_1O_2+ \angle BO_2O_1=\pi$$ so that $$ \triangle CO_2V,\ \triangle CO_1V\ (,\ \triangle O_1VO_2 )$$ are similar.
Hence $$ \triangle CO_2V : \triangle CO_1V = 16:x=x: 25 \Rightarrow x=20 $$
Hence we have $125/4$