Let \(\mathrm{C}_{1}\) be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let \(\mathrm{C}_{2}\) be the circle with centre \((1,3)\) that touches \(\mathrm{C}_{1}\) externally at the point \((\alpha, \beta)\). If \((\beta-\alpha)^{2}=\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1\), then \(\mathrm{m}+\mathrm{n}\) is equal to :

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Circle

Let

C_{1}

be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let

C_{2}

be the circle with centre

(1, 3)

that touches

C_{1}

externally at the point

(α, β)

. If

(β - α)^{2} = \frac{m}{n}, gcd (m, n) = 1

, then

m + n

is equal to :

Answer: (C)

A circle of radius

3

in the third quadrant touching both axes has centre

(- 3, - 3)

. Its distance from the centre

(1, 3)

of the second circle is

(1 + 3)^{2} + (3 + 3)^{2} = 52 = 213

. For external contact the point of tangency divides the segment joining the centres internally in the ratio

r_{1} : r_{2} = 3 : (213 - 3)

, giving

α = \frac{3 ( 1 ) + ( 2 13 - 3 ) ( - 3 )}{2 13} = \frac{12 - 6 13}{2 13}

and

β = \frac{3 ( 3 ) + ( 2 13 - 3 ) ( - 3 )}{2 13} = \frac{18 - 6 13}{2 13}

. Subtracting,

β - α = \frac{6}{2 13} = \frac{3}{13}

, so

(β - α)^{2} = \frac{9}{13}

. With

m = 9

and

n = 13

coprime,

m + n = 22

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option