Let the circle C touch the line $\mathrm{x}-\mathrm{y}+1=0$, have the centre on the positive x -axis, and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3 x+2 y=1$. Let H be the hyperbola $\frac{\mathrm{x}^{2}}{\alpha^{2}}-\frac{\mathrm{y}^{2}}{\beta^{2}}=1$, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C . Then $2 \alpha^{2}+3 \beta^{2}$ is equal to

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Circle

Let the circle C touch the line

x - y + 1 = 0

, have the centre on the positive x -axis, and cut off a chord of length

\frac{4}{13}

along the line

- 3 x + 2 y = 1

. Let H be the hyperbola

\frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1

, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C . Then

2 α^{2} + 3 β^{2}

is equal to

Answer: 19

Let the centre of

C

(p, 0)

p > 0

, with radius

r

.

Touching

x - y + 1 = 0

r = \frac{∣ p + 1∣}{2} ⟹ r^{2} = \frac{( p + 1 ) ^{2}}{2} ...(i)

Distance from the centre to

- 3 x + 2 y = 1

d = \frac{∣3 p + 1∣}{13}

. With chord length

\frac{4}{13}

r^{2} - d^{2} = (\frac{2}{13})^{2} = \frac{4}{13} ⟹ 13 r^{2} = (3 p + 1)^{2} + 4 ...(ii)

From (i) and (ii):

\frac{13 ( p + 1 ) ^{2}}{2} = (3 p + 1)^{2} + 4 ⟹ 13 p^{2} + 26 p + 13 = 18 p^{2} + 12 p + 10

5 p^{2} - 14 p - 3 = 0 ⟹ (p - 3) (5 p + 1) = 0

Since

p > 0

p = 3

, giving

r = \frac{4}{2} = 22

.

For the hyperbola

H : \frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1

, the transverse-axis length equals the diameter of

C

2 α = 2 r = 42 ⟹ α^{2} = 8

. One focus is the centre of

C

(3, 0)

, so

α e = 3

, and

α^{2} e^{2} = α^{2} + β^{2}

9 = 8 + β^{2} ⟹ β^{2} = 1

2 α^{2} + 3 β^{2} = 2 (8) + 3 (1) = 19

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Integer answer