The equation of the circle can be rewritten as : $\left ( x-5 \right )^{2}+(y-3)^{2}=25$
For finding the intersection with $x-axis$ and $y-axis$, put $x=0$ and $y=0$ in the above equation.
So we get intesection points as : $\left ( 0,3 \right )(1,0)(9,0)$
Area of the triangle formed by the line segments joining these points can be found using Heron's formula
i.e. $A=\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle defined as $s=\frac{a+b+c}{2}$ ($a,b,c$ being the sides of triangle)
For the triangle formed, $a,b,c$ can be found using Euclid's distance formula .
Then after calculating lengths of $a,b,c$, they are : $a=8,b=\sqrt{90},c=\sqrt{10}$ (values can be taken in any order ).
$\therefore$ $s=\frac{8+\sqrt{90}+\sqrt{10}}{2}=2\sqrt{10}+4$
Now, $A=\sqrt{(2\sqrt{10}+4)(2\sqrt{10}+4-8)(2\sqrt{10}+4-\sqrt{90})(2\sqrt{10}+4-\sqrt{10})}$
Simplifying, we get $A=\sqrt{(2\sqrt{10}+4)(2\sqrt{10}-4)(4-\sqrt{10})(4+\sqrt{10})}$
$\Rightarrow$ $A=\sqrt{24\times 6}=12$
Option D is correct answer.
