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? 
 A: 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
$$\left( x+\frac{13}{5} \right)^2 + \left( y +\frac{6}{5}\right)^2 = \frac{19}{5}$$
A: 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:
$$\left(x+\frac{13}{5}\right)^2+\left(y+\frac65\right)^2=\frac{19}{5}.$$
A: You can certainly keep going the way you are: solve the quadratic equation for $x$ and substitute back into the equation of the radical axis (or either circle equation) to get the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ of the chord. An equation of the circle with that diameter can be written down directly: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. Rearrange this into whatever form is required.  
However, there’s another way to solve this that requires less work, I think. The circles that pass through the two intersection points form a one-parameter family with equations that can be expressed as an affine combination of the equations of the two circles, to wit, $$(1-\lambda)(x^2+y^2+6x+2y+6)+\lambda(x^2+y^2+8x+y+10) = 0$$ or $$x^2+y^2+(2\lambda+6)x+(2-\lambda)y+(4\lambda+6)=0.\tag{*}$$ By inspection, the coordinates of the center of this circle are $(-\lambda-3,\lambda/2-1)$ and we want it to lie on the radical axis $y=2x+4$. Substitute the coordinates of the center into this equation and solve the resulting linear equation for $\lambda$, then plug that value into equation (*).
