Question

The straight line \(x + 2y = 1\) meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :

• A. \(4\sqrt{5}\)
• B. \(\frac{\sqrt{5}}{2}\)
• C. \(2\sqrt{5}\)
• D. \(\frac{\sqrt{5}}{4}\)

Hint:

AB is the diameter of the circle (angle AOB = 90 degrees); find the tangent at the origin and use the point-to-line distance formula.

Answer 1

The Correct Option is B

A = (1, 0), B = (0, 1/2). Since \(\angle AOB = 90^\circ\), AB is a diameter, so the circle is \((x-1)(x-0)+(y-0)(y-\tfrac12)=0\), i.e. \(x^2+y^2-x-\frac{y}{2}=0\).
Tangent at the origin (T = 0): \(-\frac{x}{2}-\frac{y}{4}=0 \Rightarrow 2x + y = 0\).
Distance of A(1, 0): \(\frac{|2|}{\sqrt5}=\frac{2}{\sqrt5}\). Distance of B(0, 1/2): \(\frac{1/2}{\sqrt5}=\frac{1}{2\sqrt5}\).
Sum \(= \frac{2}{\sqrt5}+\frac{1}{2\sqrt5} = \frac{5}{2\sqrt5} = \frac{\sqrt5}{2}\).
Answer: (2) \(\frac{\sqrt{5}}{2}\)