Find the sphere through the circles x^2+y^2=9, z=0 and…

Find the sphere through the circles $x^2+y^2=9,\ z=0$ and $x^2+y^2=25,\ z=2$.

Explanation of Method

The sphere through two parallel circles with a common axis.

Step 1: Center on the axis

The center of a sphere through a circle lies on the line perpendicular to the circle's plane through its center. Here that line is $Oz$, so the center is $(0,0,c)$ and

\[ x^2+y^2+(z-c)^2=R^2 . \]

Step 2: Conditions

\[ 9+c^2=R^2,\qquad 25+(2-c)^2=R^2\ \Longrightarrow\ 20-4c=0\ \Longrightarrow\ c=5,\ \ R^2=34 . \]

\[x^2+y^2+(z-5)^2=34\iff x^2+y^2+z^2-10z-9=0\]

Checks: $9-9=0$ at $z=0$, and $25+4-20-9=0$ at $z=2$. $\checkmark$

$(0,0,5)$

$x^2+y^2=9,\ z=0$
$x^2+y^2=25,\ z=2$

Both circles lie on the sphere with center $(0,0,5)$ and radius $\sqrt{34}$: $9+5^2=25+3^2=34$.

Result

\[\boxed{\,x^2+y^2+(z-5)^2=34\iff x^2+y^2+z^2-10z-9=0\,}\]