Match the column.
Normals at \(P, Q, R\) are drawn to \(y^2=4 x\) which intersect at \((3,0)\). Then,

Since, equation of normal is \(y+x t=2 a t+a t^3\) passes through \((3,0)\).
\[
\Rightarrow \quad 3 t=2 t+t^3 \Rightarrow t=0,1,-1 \text {. }
\]
\(\therefore P(1,2), Q(0,0), R(1,-2)\). Thus,
(i) Area of \(\triangle P Q R=\frac{1}{2} \times 1 \times 4=2\)
(ii) Centroid of \(\triangle P Q R=\left(\frac{2}{3}, 0\right)\)
Equation of circle passing through \(P, Q, R\) is
\[
\begin{aligned}
& \quad(x-1)(x-1)+(y-2)(y+2)+\lambda(x-1)=0 \\
& \Rightarrow \quad 1-4-\lambda=0 \Rightarrow \lambda=-3 \\
& \therefore \text { Required equation of circle is } x^2+y^2-5 x-1=0 \\
& \therefore \text { Centre }\left(\frac{5}{2}, 0\right) \text { and radius } \frac{5}{2} .
\end{aligned}
\]