<s>[INST] 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? 
 [/INST] 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:
$$\left(x+\frac{13}{5}\right)^2+\left(y+\frac65\right)^2=\frac{19}{5}.$$
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