JEE Main 2025 — 22 January, Morning Shift — Question 13

$$y = x^2 + px - 3$$ $$\text{Let } P(\alpha, 0), Q(\beta, 0), R(0, -3)$$

Circle with centre $(-1, -1)$ is $(x+1)^2 + (y+1)^2 = r^2$

Passes through $(0, -3)$

$$1^2 + (-2)^2 = r^2$$ $$r^2 = 5$$

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

Put $y=0$

$(x+1)^{2}=5-1$

$(\mathrm{x}+1)^{2}=4$

$\mathrm{x}+1= \pm 2$

$\mathrm{x}=1$ or $\mathrm{x}=-3$

$\therefore \mathrm{P}(1,0)$ and $\mathrm{Q}(-3,0)$

Area of $\triangle \mathrm{PQR}=\frac{1}{2}\left|\begin{array}{ccc}1 & 0 & 1\\ -3 & 0 & 1 \\0 & -3 & 1\end{array}\right|=6$