Example 9 The ellipse $x^{2}+2 y^{2}=2$ intersects the line $x+2 y-1=0$ at points $B$ and $C$. Point $A$ has coordinates $(2,2)$. Find the equation of the circle passing through the three points $A$, $B$, and $C$.
Solution: First, we can find the coordinates of points $B$ and $C$, and then use the method of undetermined coefficients to find the equation of the circumcircle of $\triangle ABC$.
Now, we will use the knowledge of the family of conic sections to solve it.
Since the family of conic sections $\lambda\left(x^{2}+2 y^{2}-2\right)+u(x+2 y-1)=0$ is a family of conic sections passing through points $B$ and $C$, but it does not include all curves passing through $B$ and $C$, and the circle passing through $B$ and $C$ is not included. Therefore, we need to reconstruct the family of conic sections:
$$
\lambda\left(x^{2}+2 y^{2}-2\right)+u(x+2 y-1)(x-2 y+m)=0
$$

This includes the circle passing through $B$ and $C$.
Expanding the above equation, we get:
$$
(\lambda+u) x^{2}+(2 \lambda-4 u) y^{2}+u(m-1) x+2 u(m+1) y-m u-2 \lambda=0
$$

Let $\lambda+u=2 \lambda-4 u$, and take $u=1$, we get $\lambda=5$. Substituting $\lambda=5$ into the above equation, we get:
$$
6 x^{2}+6 y^{2}+(m-1) x+2(m+1) y-m-10=0.
$$

Substituting the coordinates of point $A$ into the equation, we get $m=-8$.
Therefore, the equation that satisfies the problem is:
$$
6 x^{2}+6 y^{2}-9 x-14 y-2=0.
$$