Example 4 Find the equation of the circle which has the common chord of the intersecting circles $C_{1}: x^{2}+y^{2}+4 x+y+1=0$, and $C_{2}: x^{2}+y^{2}+2 x+2 y+1=0$ as its diameter.

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Solution 1: Subtracting the equations of the two circles yields the equation of the common chord line: $2 x-y=0$. Let the equation of the required circle be $x^{2}+y^{2}+4 x+y+1+\lambda(2 x-y)=0$, which simplifies to $x^{2}+y^{2}+(4+2 \lambda) x+(1-\lambda) y+1=0$.
The center of this circle $\left(-2-\lambda, \frac{\lambda-1}{2}\right)$ must lie on the line $2 x-y=0$, so from $2(-2-\lambda)-\frac{\lambda-1}{2}=0$, we find $\lambda=-\frac{7}{5}$.
Thus, the equation of the required circle is $5 x^{2}+5 y^{2}+16 x+12 y+5=0$.
Solution 2: The equation of the common chord of the two given circles can be found to be $2 x-y=0$.
$$
5 x^{2}+6 x+1=0 \text { and } 5 y^{2}+12 y+4=0 \text {. }
$$

Adding these two equations, we get $5 x^{2}+5 y^{2}+6 x+12 y+5=0$, which is the equation of the required circle.