# Find the equation of the sphere centered at (-9, 3, 9) with radius 5. Give an equation which describes the intersection of this sphere with the plane z = 10.

Find the equation of the sphere centered at (-9, 3, 9) with radius 5. Give an equation which describes the intersection of this sphere with the plane z = 10.
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Nichole Watt
Given: A circle with Center C = (-9,3,9) Radius r=5
General eqation for a sphere with center $C=\left({x}_{0},{y}_{0},{z}_{0}\right)$ and radius r.
$\left(x-{x}_{0}{\right)}^{2}+\left(y-{y}_{0}{\right)}^{2}+\left(z-{z}_{0}{\right)}^{2}={r}^{2}$
Replace ${x}_{0}$ with -9, ${y}_{0}$ with 3, ${z}_{0}$ with 9 and r with 5.
$\left(x-\left(-9\right){\right)}^{2}+\left(y-3{\right)}^{2}+\left(z-9{\right)}^{2}-{5}^{2}$
Simplify:
$\left(x+9{\right)}^{2}+\left(y-3{\right)}^{2}+\left(z-9{\right)}^{2}=25$
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.
$\left(x+9{\right)}^{2}+\left(y-3{\right)}^{2}+\left(10-9{\right)}^{2}=25$
Simplify:
$\left(x+9{\right)}^{2}+\left(y-3{\right)}^{2}+1=25$
Subtract 1 from each side:
$\left(x+9{\right)}^{2}+\left(y-3{\right)}^{2}=24$
CONCLUSION
Equation sphere: $\left(x+9{\right)}^{2}+\left(y-3{\right)}^{2}+\left(10-9{\right)}^{2}=25$
Equation intersection with plane z=10: $\left(x+9{\right)}^{2}+\left(y-3{\right)}^{2}=24$
###### Not exactly what you’re looking for?
Jeffrey Jordon

If a sphere has center(a, b, c) and radius r, then its equation is

$\left(x-a{\right)}^{2}+\left(y-b{\right)}^{2}+\left(z-c{\right)}^{2}={r}^{2}$

By this theorem, the equation of the sphere is:

$\left(x+9{\right)}^{2}+\left(y-3{\right)}^{3}+\left(z-9{\right)}^{2}=25$

Substitue z = 10 into the equation above, you may then get the intersection equation.