Q32 / 75 · JEE Main 2025 (22 Jan Shift 1) · Mathematics
Let the parabola $y=x^2+\mathrm{p} x-3$, 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 $\triangle P Q R$ is :
- A$7$
- B$4$
- Correct answerC$6$
- D$5$
Solution
$\mathrm{y}=\mathrm{x}^2+\mathrm{px}-3$
Let $\mathrm{P}(\alpha, 0), \mathrm{Q}(\beta, 0), \mathrm{R}(0,-3)$
Circle with centre $(-1,-1)$ is $(x+1)^2+(y+1)^2=r^2$
Passes through $(0,-3)$
$\begin{aligned}
& \left.1^2+(-2)^2=r^2\right] \\
& \mathrm{r}^2=5
\end{aligned}$
$(x+1)^2+(y+1)^2=5$
Put $\mathrm{y}=0$
$\begin{aligned}
& (x+1)^2=5-1 \\
& (x+1)^2=4 \\
& x+1= \pm 2 \\
& x=1 \text { or } x=-3 \\
& \therefore P(1,0) \text { and } Q(-3,0)
\end{aligned}$
Area of $\triangle \mathrm{PQR}=\frac{1}{2}\left|\begin{array}{ccc}1 & 0 & 1 \\ -3 & 0 & 1 \\ 0 & -3 & 1\end{array}\right|=6$
Hence, the correct answer is option: 'C'.