Find the sphere through M_1(0,0,3) and the circle…
Find the sphere through $M_1(0,0,3)$ and the circle $(x-1)^2+(y+1)^2=1,\ z=0$.
Explanation of Method
The sphere through a point and a circle in the plane z = 0.
Step 1: Pencil
Every sphere through the circle has the form
\[ (x-1)^2+(y+1)^2-1+z^2+\lambda z=0\iff x^2+y^2+z^2-2x+2y+\lambda z+1=0 . \]
Note the constant is $+1$: $1+1-1=1$.
Step 2: Through $M_1$
\[ 1+1-1+9+3\lambda=0\ \Longrightarrow\ \lambda=-\tfrac{10}3 . \]
\[3x^2+3y^2+3z^2-6x+6y-10z+3=0\]
\[ (x-1)^2+(y+1)^2+\left(z-\tfrac53\right)^2=1+\tfrac{25}9=\tfrac{34}9,\qquad \text{center }\left(1,-1,\tfrac53\right),\ \ r=\tfrac{\sqrt{34}}3 . \]
Checks: $M_1$ gives $1+1+\tfrac{16}9=\tfrac{34}9$. The circle point $(2,-1,0)$ gives $1+0+\tfrac{25}9=\tfrac{34}9$. $\checkmark$
Result
\[\boxed{\,3x^2+3y^2+3z^2-6x+6y-10z+3=0\,}\]
Remark
Correction: the source expands the pencil with constant term $-1$ instead of $+1$. Its final boxed answer $3x^2+3y^2+3z^2-6x+6y-10z-3=0$ is therefore wrong: it does not pass through $M_1$ ($27-30-3=-6\neq0$). Its radius $\tfrac{2\sqrt{13}}3$ and the "$\tfrac{34}9\neq\tfrac{52}9$" dead end come from the same slip. The correct constant is $+3$ and $r=\tfrac{\sqrt{34}}3$.