Asphere with radius 16 m is cut by a plane that is 9 m below its center. what is the best approximation of the area of the circle formed by the intersection of the plane and the sphere?

The complete question in the attached figure

we know that
the equation of a sphere is
(x-h)²+(y-k)²+(z-l)²=r²

let's assume that the center of the sphere is at the origin
so
(h,k,l)-------> (0,0,0)
for r=16 m
(x-h)²+(y-k)²+(z-l)²=r²-------->  (x)²+(y)²+(z)²=16²

for z=9 m
(x)²+(y)²+(9)²=16²--------> (x)²+(y)²=256-81--------> (x)²+(y)²=175
the radius of the circle is r=√175 m
so the area of the circle A=pi*r²-----> A=pi*175-----> A=549.5 m²

the answer is 
the area of the circle is 549.5 m²