# Find an equation of the sphere that passes through the origin and whose center is (5, -8, -10). the equation equals 0

Find an equation of the sphere that passes through the origin and whose center is (5, -8, -10). the equation equals 0
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Julianna Bell
Equation to the sphere with centre $\left({x}_{1},{y}_{1},{z}_{1}\right)$ and radius a
is $\left(x-{x}_{1}{\right)}^{2}+\left(y-{y}_{1}{\right)}^{2}+\left(z-{z}_{1}{\right)}^{2}={a}^{2}$
here $\left({x}_{1},{y}_{1},{z}_{1}\right)=\left(5,-8,-10\right)$
i.e. $\left(x-5{\right)}^{2}+\left(y+8{\right)}^{2}+\left(z+10{\right)}^{2}={a}^{2}$
and this sphere is passes through the origin so
$\left(0-5{\right)}^{2}+\left(0+8{\right)}^{2}+\left(0+10{\right)}^{2}={a}^{2}$
$⇒{a}^{2}=189$
Now required sphere equation is
$\left(x-5{\right)}^{2}+\left(y+8{\right)}^{2}+\left(z+10{\right)}^{2}=189$
$⇒{x}^{2}+{y}^{2}+{z}^{2}+189-189-10x+16y+20z=0$
$⇒{x}^{2}+{y}^{2}+{z}^{2}-10x+16y+20z=0$

pliwraih
the equation of sphere that passes through the origin can be written as
${x}^{2}+{y}^{2}+{z}^{2}={R}^{2}$ (R is the radius)
since it pass (5, -8, -10)., we have:
${5}^{2}+\left(-8{\right)}^{2}+\left(-10{\right)}^{2}={R}^{2}$
$⇒{R}^{2}=189$
the equation of sphere is
${x}^{2}+{y}^{2}+{z}^{2}=189$
$R=\sqrt{189}$ is the radius