Problem
GEO-B2-M10-P013 Two Tangent Circles
#13
★★★★☆ Level 4 of 5
Two circles touch externally at \(T\). Their common external tangent touches the circles at \(A\) and \(B\). Prove that \(\angle ATB=90^\circ\).
Choose the common tangent as the \(Ox\)-axis and put the centres at \((0,r)\) and \((d,s)\).
Let \(A=(0,0)\), \(B=(d,0)\), and let the circle centres be \(O_1=(0,r)\), \(O_2=(d,s)\). Since the circles touch externally, \(O_1O_2=r+s\), so \(d^2+(s-r)^2=(r+s)^2\), hence \(d^2=4rs\). The tangency point \(T\) divides \(O_1O_2\) in the ratio \(r:s\), so \(T=\left(\frac{rd}{r+s},\frac{2rs}{r+s}\right)\). Then \(\overrightarrow{TA}\cdot\overrightarrow{TB}=\frac{rs(4rs-d^2)}{(r+s)^2}=0\). Therefore \(TA\perp TB\).
A good problem on choosing coordinates instead of a heavy synthetic proof.