Find equation of the circle whose diameter is the common chord of two other circles?

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

$$\left( x+\frac{13}{5} \right)^2 + \left( y +\frac{6}{5}\right)^2 = \frac{19}{5}$$

You are on the right track. The numbers cancel out nicely when you sum them. Indeed: $$5x^2 + 26x + 30 = 0 \Rightarrow x_1=\frac{-13-\sqrt{19}}{5},x_2=\frac{-13+\sqrt{19}}{5}\\ y_1=\frac{-6-2\sqrt{19}}{5}, y_2=\frac{-6+2\sqrt{19}}{5}$$ The center of the new circle: $$\frac{x_1+x_2}{2}=-\frac{13}{5},\frac{y_1+y_2}{2}=-\frac{6}{5}$$ The diameter of the new circle: $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{\frac{4\cdot 19}{25}+\frac{16\cdot 19}{25}}=\sqrt{\frac{76}{5}} \Rightarrow \\ r=\frac12d=\sqrt{\frac{76}{4\cdot 5}}=\sqrt{\frac{19}{5}}$$ Thus: $$\left(x+\frac{13}{5}\right)^2+\left(y+\frac65\right)^2=\frac{19}{5}.$$