Let the parabola $y = x^2 + px - 3$, 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 $\triangle PQR$ is
Step-by-Step Solution
Key Concept: The parabola always passes through $(0,-3)$ on the $y$-axis; use the circumcircle condition at $(0,-3)$ to find the radius $r^2=5$, then find the two $x$-intercepts $P$ and $Q$ from the circle, and compute the area by the determinant formula.
$R=(0,-3)$ (y-intercept). Circle with centre $(-1,-1)$: $(x+1)^2+(y+1)^2=r^2$. Passing through $R$: $1+4=r^2=5$. Setting $y=0$: $(x+1)^2=4$, so $x=1$ or $x=-3$, giving $P=(1,0)$ and $Q=(-3,0)$. Area of $\triangle PQR = \tfrac{1}{2}\left|\det\begin{pmatrix}1&0&1\\-3&0&1\\0&-3&1\end{pmatrix}\right|=\tfrac{1}{2}|{-12}+{12}+{12}... = 6$. (Direct formula: base $|PQ|=4$, height $=|{-3}|=3$, area $=\tfrac{1}{2}\cdot4\cdot3=6$.)
Correct Answer: 3