JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Circle

A circle

C

touches the line

x = 2 y

at the point

(2, 1)

and intersects the circle

C_{1} : x^{2} + y^{2} + 2 y - 5 = 0

at two points

P

and

Q

such that

P Q

is a diameter of

C_{1}

. Then the diameter of

C

Answer: (A)

Solution

A circle touching the line

x - 2 y = 0

(2, 1)

belongs to the family

(x - 2)^{2} + (y - 1)^{2} + λ (x - 2 y) = 0 ⟹ x^{2} + y^{2} + (λ - 4) x - (2 + 2 λ) y + 5 = 0

The common chord

P Q

of this circle and

C_{1} : x^{2} + y^{2} + 2 y - 5 = 0

is their radical axis, found by subtracting:

(λ - 4) x - (4 + 2 λ) y + 10 = 0

P Q

is a diameter of

C_{1}

, so it passes through the centre

(0, - 1)

C_{1}

. Substituting gives

(4 + 2 λ) + 10 = 0

, so

λ = - 7

and

C : x^{2} + y^{2} - 11 x + 12 y + 5 = 0

Its radius is

(\frac{11}{2})^{2} + 6^{2} - 5 = \frac{245}{2}

, so the diameter of

C

245 = 75

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option