Equation of two diameters of a circle are $2 x-3 y=5$ and $3 x-4 y=7$. The line joining the points $\left(-\frac{22}{7},-4\right)$ and $\left(-\frac{1}{7}, 3\right)$ intersects the circle at only one point $\mathrm{P}(\alpha, \beta)$. Then $17 \beta-\alpha$ is equal to

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Circle

Equation of two diameters of a circle are

2 x - 3 y = 5

and

3 x - 4 y = 7

. The line joining the points

(- \frac{22}{7}, - 4)

and

(- \frac{1}{7}, 3)

intersects the circle at only one point

P (α, β)

. Then

17 β - α

is equal to

Answer: 2

Centre of circle: intersection of

2 x - 3 y = 5

and

3 x - 4 y = 7

. Solving gives

x = 1, y = - 1

, so centre

C = (1, - 1)

.

Line through

(- \frac{22}{7}, - 4)

and

(- \frac{1}{7}, 3)

: slope

= \frac{3 - ( - 4 )}{- \frac{1}{7} + \frac{22}{7}} = \frac{7}{3}

, giving

7 x - 3 y + 10 = 0 (line AB)

Since this line meets the circle at only one point

P

, it is a tangent, so

C P ⊥ A B

. The perpendicular from

C (1, - 1)

A B

3 x + 7 y + 4 = 0 (line CP)

Solving the two lines simultaneously:

7 x - 3 y = - 10, 3 x + 7 y = - 4

58 x = - 82 ⟹ x = - \frac{41}{29}, y = \frac{1}{29}

P (α, β) = (- \frac{41}{29}, \frac{1}{29})

17 β - α = 17 (\frac{1}{29}) - (- \frac{41}{29}) = \frac{17 + 41}{29} = \frac{58}{29} = 2

Key confirmed: 2.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer