Find equation of the circle whose diameter is the common chord of two other circles? Circle 1:
$$x^2 + y^2 +6x + 2y +6 = 0$$
Circle 2:
$$x^2 + y^2 + 8x + y + 10 = 0$$
My attampt:
From circle 1 and 2, I found 
$$ y = 2x + 4 $$
which is the common chord. 
Pluging that in equation 1 I got
$$5x^2 + 26x + 30 = 0$$
here I got stuck for too much complication if I use quadratic formula.Is there anything wrong? 

First, obtain the equations of the intersection points below for both $x$ and $y$,
$$5x^2 + 26x + 30= 0$$
$$5y^2 + 12y -8= 0$$
It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,
$$x_1+x_2=-\frac{26}{5},\>\>\>x_1x_2=6$$
$$y_1+y_2=-\frac{12}{5},\>\>\>y_1y_2=-\frac 85 $$
Thus, the center of the circle is $\frac{x_1+x_2}{2}=-\frac{13}{5}, \frac{y_1+y_2}{2}=-\frac{6}{5}$ and its diameter squared is,
$$(x_1-x_2)^2 + (y_1-y_2)^2$$
$$ = (x_1+x_2)^2- 4x_1x_2 + (y_1+y_2)^2 - 4y_1y_2$$
$$= \left( \frac{26}{5} \right)^2 -4\cdot 6 + \left( \frac{12}{5}\right)^2 + 4\cdot \frac 85 = \frac{76}{5}$$
The equation of the circle is
$$\left( x+\frac{13}{5} \right)^2 + \left( y +\frac{6}{5}\right)^2 = \frac{19}{5}$$
