# Find an equation for each sphere that passes through the point(5,1,4) and is tangent to all three coordinate planes.

Find an equation for each sphere that passes through the point(5,1,4) and is tangent to all three coordinate planes.
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kunstboom8w
the general sphere equation is
$\left(x-{x}_{0}{\right)}^{2}+\left(y-{y}_{0}{\right)}^{2}+\left(z-{z}_{0}{\right)}^{2}={R}^{2}$...(1)
$\left({x}_{0},{y}_{0},{z}_{0}\right)$ is the centerof sphere
since the sphere tangent to all three coordinate planes, we have:
${x}_{0}={y}_{0}={z}_{0}=R$
(1) change to:
$\left(x-R{\right)}^{2}+\left(y-R{\right)}^{2}+\left(z-R{\right)}^{2}=R$
sphere passes through the point (5,1,4) , we have:
$\left(5-R{\right)}^{2}+\left(1-R{\right)}^{2}+\left(4-R{\right)}^{2}=R$
$⇒25-10R+{R}^{2}+1-2R+{R}^{2}+16-8R+{R}^{2}={R}^{2}$
$2{R}^{2}-20R+32=0$
${R}^{2}-10R+16=0$
(R-2)(R-8) = 0
R=2, or R=8
the sphere equations are
$\left(x-2{\right)}^{2}+\left(y-2{\right)}^{2}+\left(z-2{\right)}^{2}={2}^{2}$
or
$\left(x-8{\right)}^{2}+\left(y-8{\right)}^{2}+\left(z-8{\right)}^{2}={8}^{2}$
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popljuvao69
everything answered above is totally perfect except that 25+1+16=42, and not 25+1+16=32. So, radius equals 3 and 7.