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

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Circle

The straight line

x + 2 y = 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) $\frac{5}{2}$ ✓ Correct
(B) $25$
(C) $\frac{5}{4}$
(D) $45$

Answer: (A)

Solution

The line

x + 2 y = 1

cuts the axes at

A (1, 0)

and

B (0, \frac{1}{2})

, and the circle through

O

A

and

B

x^{2} + y^{2} - x - \frac{y}{2} = 0

. The tangent at the origin is obtained by keeping the first-degree terms,

- x - \frac{y}{2} = 0

, i.e.

2 x + y = 0

. The distance from

A

\frac{∣2∣}{5} = \frac{2}{5}

and from

B

\frac{∣1/2∣}{5} = \frac{1}{2 5}

, and their sum is

\frac{2}{5} + \frac{1}{2 5} = \frac{5}{2 5} = \frac{5}{2}

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

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