Circles
15.0 Common chord of two circles
15.0 Common chord of two circles
The chord joining the points of intersection of two given circles is called their common chord. In figure $45$, $PQ$ is the 
common chord of two intersecting circles $$S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0$$ and $$S' \equiv {x^2} + {y^2} + 2g'x + 2f'y + c' = 0$$
common chord of two intersecting circles $$S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0$$ and $$S' \equiv {x^2} + {y^2} + 2g'x + 2f'y + c' = 0$$Then the equation of common chord is $S - S{\text{'}} = 0$ i.e., $$2x\left( {g - g'} \right) + 2y\left( {f - f'} \right) + c - c' = 0$$
Proof: Since $S=0$ and $S'=0$ are two intersecting circles, then $S-S'=0$
or, $$2x\left( {g - g'} \right) + 2y\left( {f - f'} \right) + c - c' = 0$$
is a first degree equation in $x$ and $y$ which represents a straight line and equation satisfies the intersecting points $P$ and $Q$ of two given circles $S=0$ and $S'=0$.
