Question

A sphere 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?

Answer 1

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