The two circles below are externally tangent. A common external tangent intersects line PQ at R. Find QR.
+1768 Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.
Since PQ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.
Let x=QR. Since ∠ROQ is a straight angle,
[\angle TPO + \angle TQA = 180^\circ,]
so ∠TPO=∠TQA=90∘.
Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]
Then 4⋅14=x⋅16+64, so x=2.
