Consider a circle with centre be $\left(a,b\right)$.

So the equation of circle will be,

$x^2+y^2-2ax-2by+c=0 \cdots \left(1\right)$

In the question it is given that the circle passes through $\left(2,3\right)$.

Therefore, substituting $x=2$ and $y=3$ respectively in the equation $\left(1\right)$ we get,

$2^2+3^2-2a\times 2-2b\times 3+c=0$

$\Rightarrow 13-4a-6b+c=0$

$\therefore c=4a+6b-13$

So now putting the value of $c$ in the equation $\left(1\right)$ we get,

$x^2+y^2-2ax-2by+\left(4a+6b-13\right)=0 \cdots \left(2\right)$

Now we will find intercept on line $x=2, y=3$ respectivley.

By substituting $x=2$ in the equation $\left(1\right)$ we get,

$4+y^2-4a-2by+\left(4a+6b-13\right)=0 \cdots \left(3\right)$

In the equation $\left(3\right)$ let $y_1,y_2$ be the roots.

So, from equation $\left(3\right)$ sum of the roots will be,

$y_1+y_2=2b \dots \left(4\right)$

Product of the roots will be,

$y_1{y}_2=6b-9 \cdots \left(5\right)$

Now as we know that length of intercept is $\left(y_1-y_2\right)$. So we will use the identity, $a-b=\sqrt{{\left(a+b\right)}^2-4ab}$

$\Rightarrow y_1-y_2=\sqrt{{\left(y_1+y_2\right)}^2-4y_1{y}_2}$

$\Rightarrow 3^2={\left(y_1+y_2\right)}^2-4y_1{y}_2$

Substituting the values of equation $\left(4\right)$ and $\left(5\right)$, we get

$\Rightarrow 9={\left(2b\right)}^2-4\left(6b-9\right)$

$\Rightarrow 4b^2-24b+27=0$

$\Rightarrow 4b^2-6b-18b+27=0$

$\Rightarrow 2b\left(2b-3\right)-9\left(2b-3\right)=0$

$\Rightarrow \left(2b-9\right)\left(2b-3\right)=0$

$\Rightarrow b=\frac{9}{2},b=\frac{3}{2}$

Hence the values of $b$ is $\frac{9}{2}$ or $\frac{3}{2}$

By substituting the value of $y=3$ in the euqation $\left(2\right)$ we get,

$x^2+9-2ax-6b+\left(4a+6b-13\right)=0$

$x^2-2ax+4a-4=0 \cdots \left(6\right)$

In equation $\left(6\right)$ let $x_1,x_2$ be the roots,

Sum of the roots will be,

$x_1+x_2=2a \cdots \left(7\right)$

Product of the roots will be

$x_1{x}_2=\left(4a-4\right) \cdots \left(8\right)$

Again using same identity we get,

$\left(x_1-x_2\right)=\sqrt{{\left(x_1+x_2\right)}^2-4x_1{x}_2}$

$\Rightarrow 4={\left(x_1+x_2\right)}^2-4x_1{x}_2$

$\Rightarrow 4={\left(2a\right)}^2-4\left(4a-4\right)$

$\Rightarrow 4=a^2-\left(4a-4\right)$

$\Rightarrow a^2-4a=0$

$\Rightarrow a\left(a-4\right)=0$

Here the value of $a$ is $0$ or $4$.

So now there are $4$ such sets of equation can be made by the sets of vales of $\left(a,b\right)$ which are

$\left(0,\frac{3}{2}\right),\left(0,\frac{9}{2}\right),\left(4,\frac{3}{2}\right)$ and $\left(4,\frac{9}{2}\right)$.

For $\left(0,\frac{3}{2}\right)$ we will substitute in equation $\left(3\right)$, we get

$x^2+y^2-2ax-2by+\left(4a+6b-13\right)=0$

$x^2+y^2-2x\left(0\right)-2y\times \frac{3}{2}+\left(4\left(0\right)+6\times \frac{3}{2}-13\right)=0$

$\therefore x^2+y^2-3y+\left(-4\right)=0$

For $\left(0,\frac{9}{2}\right)$ we will substitue in equation $\left(3\right)$, we get

$x^2+y^2-2ax-2by+\left(4a+6b-13\right)=0$

$x^2+y^2-2x\left(0\right)-2y\times \frac{9}{2}+\left(4\left(0\right)+6\times \frac{9}{2}-13\right)=0$

$\therefore x^2+y^2-9y+14=0$

For $\left(4,\frac{3}{2}\right)$ we will substitue in euqation $\left(3\right)$, we get

$x^2+y^2-2ax-2by+\left(4a+6b-13\right)=0$

$x^2+y^2-2x\left(4\right)-2y\times \frac{3}{2}+\left(4\left(4\right)+6\times \frac{3}{2}-13\right)=0$

$\therefore x^2+y^2-8x-3y+12=0$

For $\left(4,\frac{9}{2}\right)$ we will substitute in equation $\left(3\right)$, we get

$x^2+y^2-2ax-2by+\left(4a+6b-13\right)=0$

$x^2+y^2-2x\left(4\right)-2y\times \frac{9}{2}+\left(4\left(4\right)+6\times \frac{9}{2}-13\right)=0$

$\therefore x^2+y^2-8x-9y+30=0$

So only $x^2+y^2-8x-9y+30=0$ is matching with the given option.