Question

Let \(\mathrm{PL} = 8\) units and \(\mathrm{QM} = 2\) units be two parallel line segments such that the line segments \(\mathrm{PM}\) and \(\mathrm{QL}\) intersect at the point \(\mathrm{R}\). If \(\mathrm{PL}\) and \(\mathrm{QM}\) are tangents to a circle passing through points \(\mathrm{P}, \mathrm{Q}, \mathrm{R}\) then radius of this circle is

• A. \(\sqrt{2}\)
• B. 2
• C. \(2\sqrt{2}\)
• D. 4

Hint:

The circle touches both parallel lines at P and Q, so PQ is a diameter perpendicular to the lines; place coordinates and use that R lies on the circle.

Answer 1

The Correct Option is B

The circle passes through P (on line PL) and is tangent to line PL, so it touches PL at P; similarly it touches line QM at Q. Tangents at P and Q are parallel, so PQ is a diameter and PQ is perpendicular to both lines. Let PQ = d.
Take P = (0, 0), Q = (d, 0), L = (0, 8), M = (d, 2).
Line PM: \((x, y) = s(d, 2)\). Line QL: \((x, y) = (d, 0) + u(-d, 8)\). Equating: \(sd = d - ud\) and \(2s = 8u\), so \(s = 0.8\), \(u = 0.2\). Hence R = (0.8d, 1.6).
The circle has centre (d/2, 0) and radius d/2. R lies on it: \((0.3d)^2 + 1.6^2 = (d/2)^2\), i.e. \(0.09d^2 + 2.56 = 0.25d^2\), so \(0.16d^2 = 2.56\), \(d^2 = 16\), \(d = 4\).
Radius = d/2 = 2.
Answer: (2) 2