Find the circle whose diameter is the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$.

Solution:

Given circle equations are $S\equiv x^2+y^2+2x+3y+1=0$ and $S\text{'}\equiv x^2+y^2+2x+3y+1=0$.

The formula for common chord of two circles is $S-S^{\text{'}}=0$

$$\begin{gathered} \Rightarrow S-S\text{'}=x^2+y^2+2x+3y+1-(x^2+y^2+4x+3y+2) \\ =2x-4x+1-2=-2x-1 \end{gathered}$$ 

The equation of common chord is $L\equiv 2x+1=0.$

Now, the required circle must pass through the point of intersection of $S$ and $S^{\text{'}}$ to have their common chord as diameter.

Let the required circle is $S\text{'}\text{'}\equiv x^2+y^2+ax+by+c=0$.

The equation of any circle passing through the intersection of $S,S\text{'}$ is $S+\lambda L.$

$$\begin{gathered} \Rightarrow S\text{'}\text{'}\equiv S+\lambda L=x^2+y^2+2x+3y+1+\lambda \left(2x+1\right)=0 \\ =x^2+y^2+\left(2+2\lambda \right)x+3y+1+\lambda =0 \end{gathered}$$

The center of a circle $x^2+y^2+ax+by+c=0$ is $\left(\frac{-a}{2},\frac{-b}{2}\right)$ and it's radius is $r=\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2-c}$ .

$\Rightarrow$Center of $S\text{'}\text{'}$ is $C\left(-\left(\lambda +1\right),\frac{-3}{2}\right)$.

As the diameter passes through the center of the circle i.e. $2x+1$ passes through the center.

$$\begin{gathered} \Rightarrow 2\left(-\lambda -1\right)+1=0=-2\lambda -2+1 \\ \Rightarrow \lambda =-\frac{1}{2}. \end{gathered}$$

Then the equation of the circle will be,

$$\begin{gathered} S\text{'}\text{'}\equiv x^2+y^2+2\left(1-\frac{1}{2}\right)x+3y+1-\frac{1}{2}=0 \\ \equiv x^2+y^2+x+3y+\frac{1}{2}=0 \end{gathered}$$

Therefore, the equation of the required circle is $S\text{'}\text{'}\equiv 2x^2+2y^2+2x+6y+1=0$.

Therefore, the correct answer is option $1.$