Radical axis $P Q$ is $x-\frac{a}{2}=0$ which cuts the circle $x^2+y^2=a^2$ in $P$ and $Q$. Now, any circle which passes through the points $P$ and $Q$ will have radical axis the line $P Q$ with respect to $x^2+y^2-a^2=0$. Hence, its equation is the equation of the circle through the points of intersection of the circle.
$$
x^2+y^2-a^2=0
$$


and the line $x-\frac{a}{2}=0$ and is given by $s+\lambda p=0$
i.e. $\left(x^2+y^2-a^2\right)+\lambda\left(x-\frac{a}{2}\right)=0$
As, it passes through the point $(2 a, 0)$.
$$
\therefore \quad\left(4 a^2-a^2\right)+\lambda\left(2 a-\frac{a}{2}\right)=0 \Rightarrow \lambda=-2 a
$$
Hence, the equation of circle $C$ is
$$
\begin{aligned}
& \left(x^2+y^2-a^2\right)-2 a\left(x-\frac{a}{2}\right)=0 \\
\Rightarrow & x^2+y^2-2 a x=0 \quad \Rightarrow(x-a)^2+y^2=a^2
\end{aligned}
$$
Whose centre is $(a, 0)$ and passes through $(0,0)$ and $(a, a)$.