JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Straight Lines

Let a variable line passing through the centre of the circle

x^{2} + y^{2} - 16 x - 4 y = 0

, meet the positive co-ordinate axes at the point A and B. Then the minimum value of

OA + OB

, where O is the origin, is equal to

Answer: (B)

Solution

The circle

x^{2} + y^{2} - 16 x - 4 y = 0

has centre

(8, 2)

.

A line through

(8, 2)

can be written as

y - 2 = m (x - 8)

.

x-intercept (A): put

y = 0

x = 8 - \frac{2}{m}

.
y-intercept (B): put

x = 0

y = 2 - 8 m

.

For both intercepts to be positive,

m

must be negative. Write

m = - k

k > 0

O A + O B = (8 + \frac{2}{k}) + (2 + 8 k) = 10 + \frac{2}{k} + 8 k

.

By AM–GM,

\frac{2}{k} + 8 k \geq 2 \frac{2}{k} \cdot 8 k = 216 = 8

, with equality when

\frac{2}{k} = 8 k

, i.e.

k^{2} = \frac{1}{4}

k = \frac{1}{2}

.

Minimum value of

O A + O B = 10 + 8 = 18

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option