Ellipse would have been easier

The most likely place for the circle is on the left side of the diagram. It will touch the bottom (what I will call the x axis), the left side (what I will call the y axis) and the corner at the top of the diagram which protrudes down into it (which becomes point $P$ $(8,9)$ on our coordinate plane).

Because it's a circle, every point on its perimeter will be an equal distance $r$ from its center. This means that the $(x,y)$ coordinates of the center will simply be $(r,r)$ , and point $P$ will be also be a distance $r$ from the center. So, using the distance formula:

$x=y=r=\sqrt{(r-x_{P})^2+(r-y_{P})^2}$

$r=\sqrt{(r-8)^2+(r-9)^2}$

Square both sides:

$r^2=(r-8)^2+(r-9)^2$

$0=r^2-34r+145$

$0=(r-29)(r-5)$

$r$ can't be 29 because of the size of the diagram, so it is 5.

$A=πr^2=25π$

Let left bottom corner be (0,0) and x-axis along 18 side, y-axis along 12 side.

$16\pi$ does not even fill the right smaller space. It is too small.
$\dfrac{145}{4}*\pi$ goes out.
$\dfrac{81} 4*\pi$ fits the small right portion. If there is no other bigger option this would be the solution.
$25*\pi$ has a radius of 5. So the center of the circle is (5,?). The horizontal semi-chord of the circle at the corner (8,9) must be 8 - 5 = 3 wide. So the distance of the semi chord from the center =\sqrt{5^2 - 3^2)=4. Distance available is also 9 - 5=4.So the circle just touches the corner. No circle bigger than this is possible.