Let the line $\mathrm{x}-\mathrm{y}=4$ intersect the circle $\mathrm{C}:(\mathrm{x}-4)^{2}+(\mathrm{y}+3)^{2}=9$ at the points Q and R . If $\mathrm{P}(\alpha, \beta)$ is a point on C such that $\mathrm{PQ}=\mathrm{PR}$, then $(6 \alpha+8 \beta)^{2}$ is equal to

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Circle

Let the line

x - y = 4

intersect the circle

C : (x - 4)^{2} + (y + 3)^{2} = 9

at the points Q and R . If

P (α, β)

is a point on C such that

PQ = PR

, then

(6 α + 8 β)^{2}

is equal to

Answer: 18

Solution

Putting

x = y + 4

in the circle,

y^{2} + (y + 3)^{2} = 9

, i.e.

2 y^{2} + 6 y = 0

, so

y = 0

y = - 3

; the intersection points are

Q (4, 0)

and

R (1, - 3)

. The condition

P Q = P R

places

P

on the perpendicular bisector of

QR

, which passes through the centre

(4, - 3)

. Since

QR

has slope

1

, this bisector has slope

- 1

and is

x + y = 1

, so

α + β = 1

. With

β = 1 - α

on the circle,

(α - 4)^{2} + (4 - α)^{2} = 9

, giving

(α - 4)^{2} = \frac{9}{2}

and

α = 4 \pm \frac{3}{2}

. Then

6 α + 8 β = 6 α + 8 (1 - α) = 8 - 2 α = \mp 32

, so

(6 α + 8 β)^{2} = 18

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Integer answer

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