In the given figure, two identical circles of radius 5 cm with centres A and B touch each other, and Q, A, B, P lie on one line, with Q and P on the two circles. RQ touches the circle with centre A at Q, and RP touches the same circle at S. What is the length of RQ (in cm)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(5\sqrt{2}\)
A tangent is perpendicular to the radius at the point of contact, so \(\angle ASP=90^\circ\) and \(\angle RQP=90^\circ\). Triangles PSA and PQR then share \(\angle P\) and are similar.
\(PQ=4\times5=20\,\mathrm{cm}\), \(PA=PB+BA=5+10=15\,\mathrm{cm}\)
\(PS=\sqrt{PA^{2}-AS^{2}}=\sqrt{225-25}=\sqrt{200}=10\sqrt{2}\,\mathrm{cm}\)
Similar triangles: \(\displaystyle \frac{RQ}{AS}=\frac{PQ}{PS}\)
\(\displaystyle RQ=\frac{5\times20}{10\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}\,\mathrm{cm}\)
