Given, equation of line, $2 x+3 y-k=0, k>0$.
$\frac{x}{\frac{k}{2}}+\frac{y}{\frac{k}{3}}=1 \quad \text{...(i)}$



Given area of $\triangle \mathrm{OAB}=12$
$$
\begin{gathered}
=\frac{1}{2}\left|\begin{array}{ccc}
0 & 0 & 1 \\
\frac{\mathrm{k}}{2} & 0 & 1 \\
0 & \frac{\mathrm{k}}{3} & 1
\end{array}\right|=12 \\
\Rightarrow \quad \frac{\mathrm{k}^{2}}{6}=24 \Rightarrow \mathrm{k}^{2}=144 \Rightarrow \mathrm{k}=12
\end{gathered}
$$
Hence, the coordinate of $A(6,0)$ and $B(0,4)$.
Then, diameter $=\mathrm{AB}$
$$
\begin{aligned}
&=\sqrt{(6-0)^{2}+(0-4)^{2}} \\
&=\sqrt{36+16} \\
&=\sqrt{52}
\end{aligned}
$$
Centre $=$ Mid point of $A B$
$$
=\left(\frac{6+0}{2}, \frac{0+4}{2}\right)=(3,2)
$$
The equation of circle is
$\Rightarrow \quad(x-3)^{2}+(y-2)^{2}=\left(\frac{\sqrt{52}}{2}\right)^{2}$
$\Rightarrow \mathrm{x}^{2}+9-6 \mathrm{x}+\mathrm{y}^{2}+4-4 \mathrm{y}=13$
$\Rightarrow \quad x^{2}+y^{2}-6 x-4 y+13=13$
$\Rightarrow \quad x^{2}+y^{2}-6 x-4 y=0$