JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Differential Equations

Let

\frac{d y}{d x} = \frac{a x - b y + a}{b x + cy + a}

, where

a, b, c

are constants, represent a circle passing through the point

(2, 5)

. Then the shortest distance of the point

(11, 6)

from this circle is :

Answer: (B)

Solution

For a circle

x^{2} + y^{2} + 2 g x + 2 f y + k = 0

, the differential equation is:

\frac{d y}{d x} = - \frac{2 x + 2 g}{2 y + 2 f} = - \frac{x + g}{y + f}

Comparing with

\frac{d y}{d x} = \frac{a x - b y + a}{b x + cy + a}

, we need

b = 0

a = - 2

c = 2

, giving

g = 1

and

f = - 1

.

The circle equation becomes

x^{2} + y^{2} + 2 x - 2 y + k = 0

. Passing through

(2, 5)

4 + 25 + 4 - 10 + k = 0 \Rightarrow k = - 23

So the circle is

x^{2} + y^{2} + 2 x - 2 y - 23 = 0

with center

C = (- 1, 1)

and radius

r = 1 + 1 + 23 = 5

.

Distance from

P = (11, 6)

to center:

d = (11 + 1)^{2} + (6 - 1)^{2} = 144 + 25 = 13

.

Shortest distance from

P

to circle:

d - r = 13 - 5 = 8

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option