How to read the equation of a sphere and identify its center and radius

Key takeaway: If you see a surface written in standard sphere form, the key is to match each term with the center shift and the constant with the radius. The notation here is already doing most of the work for you.

Write the equation in standard form:

$x^2+y^2+(z-2)^2=8$

This matches

$(x-h)^2+(y-k)^2+(z-l)^2=r^2$

so:

• center = $(0,0,2)$
• radius = $\sqrt{8}=2\sqrt{2}$

The condition $z\ge 0$ means we only keep the part of the sphere above the plane $z=0$.

So yes, it is not centered at the origin. The center is shifted up to $(0,0,2)$, and the surface is a sphere of radius $\sqrt{8}$.

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Pitfalls the pros know 👇 A common slip is to read $x^2+y^2+(z-2)^2=8$ as centered at $(0,0,0)$ just because the $x$ and $y$ terms are unchanged. The shift in $z$ is the important part. Another easy mistake is forgetting that the right-hand side is $r^2$, not $r$.

What if the problem changes? If the equation were $x^2+y^2+z^2=8$, then the center would be the origin and the radius would still be $\sqrt{8}$. If the equation were $x^2+y^2+(z+2)^2=8$, the center would be $(0,0,-2)$ instead.

Tags: sphere equation, center and radius, standard form