 # Find an equation for each sphere that passes through the point(5,1,4) and is tangent to all three coordinate planes. Taliyah Reyes 2022-08-03 Answered
Find an equation for each sphere that passes through the point(5,1,4) and is tangent to all three coordinate planes.
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it 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
$\left(x-8{\right)}^{2}+\left(y-8{\right)}^{2}+\left(z-8{\right)}^{2}={8}^{2}$
###### Not exactly what you’re looking for? 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.