The equation of the circle whose diameter is the common chord of the circles  $x^2+y^2+2x+2y+1=0\text{ and }{x}^2+y^2+4x+6y+4=0$ is

Equation of the common chord is $2x+4y+3=0$.

Equation of the circle passing through the points of intersection of the given circles is 

$x^2+y^2+2x+2y+1+\lambda (2x+4y+3)=0\Rightarrow x^2+y^2+(2+2\lambda )x+(2+4\lambda )y+(1+3\lambda )=0\to \left(1\right)$

Centre of (1)  is $(-1-\lambda ,-1-2\lambda )$.

IF common chord is a diameter of (1), then $2(-1-\lambda )+4(-1-2\lambda )+3=0$

$\Rightarrow -2-2\lambda -4-8\lambda +3=0\Rightarrow 10\lambda =-3\Rightarrow \lambda =-3/10$.

Equation of the required circle is $x^2+y^2+(2-3/5)x+(2-6/5)y+(1-9/10)=0$

$\Rightarrow 10x^2+10y^2+14x+8y+1=0$.