There are two concentric circles with centre P. The larger circle has radius PQ = 10 cm, and the smaller circle has radius PR = 8 cm. TQ is a tangent to the larger circle from an external point T (touching it at Q), and TR is a tangent to the smaller circle from the same point T (touching it at R). If TQ = 24 cm, find the value of (TR – PR)/PQ.
\(\frac{3\sqrt{17} - 4}{5}\)
A tangent is perpendicular to the radius at the point of contact, so triangles TQP and TRP are right-angled at Q and R.
In triangle TQP: \(TP^{2} = TQ^{2} + PQ^{2} = 24^{2} + 10^{2} = 576 + 100 = 676\), so TP = 26 cm.
In triangle TRP: \(TR^{2} = TP^{2} - PR^{2} = 676 - 64 = 612\), so \(TR = \sqrt{612} = 6\sqrt{17}\) cm.
Then \(\frac{TR - PR}{PQ} = \frac{6\sqrt{17} - 8}{10} = \frac{3\sqrt{17} - 4}{5}\).
Check: \((6\sqrt{17})^{2} + 8^{2} = 612 + 64 = 676 = 26^{2}\), so TR and PR fit the same TP = 26 cm found from the larger circle.
Hence, \(\frac{3\sqrt{17} - 4}{5}\).
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