Given: A circle with Center C = (-9,3,9) Radius r=5

General eqation for a sphere with center \(C=(x_0,y_0,z_0)\) and radius r.

\((x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2\)

Replace \(x_0\) with -9, \(y_0\) with 3, \(z_0\) with 9 and r with 5.

\((x-(-9))^2+(y-3)^2+(z-9)^2-5^2\)

Simplify:

\((x+9)^2+(y-3)^2+(z-9)^2=25\)

The intersection of the sphere with the plane z=10 can then be found by replacing z in the equation of the sphere by 10.

\((x+9)^2+(y-3)^2+(10-9)^2=25\)

Simplify:

\((x+9)^2+(y-3)^2+1=25\)

Subtract 1 from each side:

\((x+9)^2+(y-3)^2=24\)

CONCLUSION

Equation sphere: \((x+9)^2+(y-3)^2+(10-9)^2=25\)

Equation intersection with plane z=10: \((x+9)^2+(y-3)^2=24\)

General eqation for a sphere with center \(C=(x_0,y_0,z_0)\) and radius r.

\((x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2\)

Replace \(x_0\) with -9, \(y_0\) with 3, \(z_0\) with 9 and r with 5.

\((x-(-9))^2+(y-3)^2+(z-9)^2-5^2\)

Simplify:

\((x+9)^2+(y-3)^2+(z-9)^2=25\)

The intersection of the sphere with the plane z=10 can then be found by replacing z in the equation of the sphere by 10.

\((x+9)^2+(y-3)^2+(10-9)^2=25\)

Simplify:

\((x+9)^2+(y-3)^2+1=25\)

Subtract 1 from each side:

\((x+9)^2+(y-3)^2=24\)

CONCLUSION

Equation sphere: \((x+9)^2+(y-3)^2+(10-9)^2=25\)

Equation intersection with plane z=10: \((x+9)^2+(y-3)^2=24\)