The two circles below are externally tangent. A common external tangent intersects line \(PQ\) at \(R\) Find \(QR\).
Guest, please stop posting random answers.
We know PQ is 18, and by pythagorean theorem, we can conclude that RS (the points of tangency of circles P and Q respectively) has length sqrt(18^2 - 2^2). sqrt(320) simplifies to 8sqrt(5). (unnessecary but you can do it this way)
Let QR = x. By similar triangles, x/8 = (x + PQ)/10
10x = 8x + 8*18
2x = 144
x = 72 = QR