What would this make 'Rhombus with in two circle'?

It helps if you sketch the diagram.

${ x }^{ 2 }-4x-12=-{ y }^{ 2 }\\ { x }^{ 2 }+4x-12=-{ y }^{ 2 }$

Observe that the equations/circles will meet when $x=0$ which gives $y=\pm \sqrt { 12 }$ . Plot these two points.

Since these two points are reflection of each others along the x-axis, the circles are symmetrical along the x-axis.

Then, finding the roots of the equations: ${ x }^{ 2 }-4x-12=0\\ { x }^{ 2 }+4x-12=0$

For the first equation we get $x=6,-2, y=0$ as points on the circle.

For the second equation we get $x=2,-6, y=0$ as points on the circle.

Thus, the points on the overlapping region is (-2,0) and (2,0). Area of rhombus: $[2-(-2)]\times [\sqrt { 12 } -(-\sqrt { 12 } )]\div 2=4\sqrt { 12 } =8\sqrt { 3 }$

8+3= $\boxed{11}$ .