Let the parabola $y=x^{2}+px-3$, meet the coordinate axes at the points $\mathrm{P}, \mathrm{Q}$ and R. If the circle C with centre at $(-1,-1)$ passes through the points $P, Q$ and $R$, then the area of $\triangle PQR$ is :

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Circle

Let the parabola

y = x^{2} + p x - 3

, meet the coordinate axes at the points

P, Q

and R. If the circle C with centre at

(- 1, - 1)

passes through the points

P, Q

and

R

, then the area of

△ P QR

is :

Answer: (B)

The parabola

y = x^{2} + p x - 3

crosses the

y

-axis at

x = 0

, giving the point

R (0, - 3)

.

The circle centered at

(- 1, - 1)

passes through

R

, so its radius satisfies

r^{2} = (0 - (- 1))^{2} + (- 3 - (- 1))^{2} = 1 + 4 = 5.

Circle:

(x + 1)^{2} + (y + 1)^{2} = 5

. The parabola meets the

x

-axis (

y = 0

) at points

P, Q

, and these too lie on the circle, so setting

y = 0

in the circle's equation:

(x + 1)^{2} + 1 = 5 ⟹ (x + 1)^{2} = 4 ⟹ x + 1 = \pm 2 ⟹ x = 1 or x = - 3.

P (1, 0)

and

Q (- 3, 0)

.

Area of

△ P QR

with

P (1, 0), Q (- 3, 0), R (0, - 3)

Area = \frac{1}{2} ∣ x_{P} (y_{Q} - y_{R}) + x_{Q} (y_{R} - y_{P}) + x_{R} (y_{P} - y_{Q}) ∣ = \frac{1}{2} ∣ 1 (0 + 3) + (- 3) (- 3 - 0) + 0 ∣ = \frac{1}{2} (3 + 9) = 6.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option