Example 4. Find the circle intersecting the given circle $x^{2}+y^{2}-7 y+10$ $=0$ such that the common chord is parallel to the given line $2 \mathbf{x}$ $-3 y-1=0$, and passes through the points $(-2,3)$ and $(1,4)$.
The equation of the circle passing through the points $(-2,3),(1,4)$ is
$$
\begin{aligned}
& (x+2)(x-1)+(y-3)(y-4) \\
+\lambda & {[(x+2)-(y-3) \cdot 3]=0 }
\end{aligned}
$$

which simplifies to \(x^{2}+y^{2}+(\lambda+1) x-(3 \lambda+7) y+11 \lambda\)
$$
+10=0 \text {. }
$$

Since (·) intersects with the known circle, subtracting the two equations gives the equation of the common chord as
$$
(\lambda+1) x-3 \lambda y+11 \lambda=0 .
$$

Furthermore, since (·)′ is parallel to the line \(2 x-3 y-1=0\),
we have
$$
\frac{\lambda+1}{2}=\frac{-3 \lambda}{-3}, \lambda=1 .
$$

Substituting \(\lambda=1\) into (·) yields the equation of the required circle as
$$
x^{2}+y^{2}+2 x-10 y+21=0
$$