JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Circle

If one of the diameters of the circle

x^{2} + y^{2} - 10 x + 4 y + 13 = 0

is a chord of another circle

C

, whose center is the point of intersection of the lines

2 x + 3 y = 12

and

3 x - 2 y = 5

, then the radius of the circle C is

Answer: (C)

Solution

The circle

x^{2} + y^{2} - 10 x + 4 y + 13 = 0

has centre

(5, - 2)

and radius

25 + 4 - 13 = 16 = 4

.

The centre of circle

C

is the intersection of

2 x + 3 y = 12

and

3 x - 2 y = 5

. Solving: from the two equations,

13 x = 39 \Rightarrow x = 3

, and then

y = 2

. So the centre of

C

(3, 2)

.

A diameter of the first circle is a chord of

C

, so the distance from

C

's centre

(3, 2)

to the first circle's centre

(5, - 2)

, call it

d

, together with the first circle's radius

r_{1} = 4

(half the chord), satisfies

R^{2} = d^{2} + r_{1}^{2}

d = (5 - 3)^{2} + (- 2 - 2)^{2} = 4 + 16 = 20

R = d^{2} + r_{1}^{2} = 20 + 16 = 36 = 6

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option