Common Chord of two Circles - Practice Questions & MCQ

Common Chord of Two Circles

If two circles $S=0$ and $S^{\prime}=0$, intersects at two points let say at $A$ and $B$. Then, the equation of the line joining the points $A$ and $B$ is called the common chord of the two circles.
The equation of the common chord of two circles

$
\begin{aligned}
& (\mathrm{S})=\mathrm{x}^2+\mathrm{y}^2+2 g \mathrm{~g}+2 \mathrm{fy}+\mathrm{c}=0 \\
& \left(S^{\prime}\right)=x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0 \text { is } \\
& \mathbf{2 x}\left(\mathbf{g}-\mathbf{g}^{\prime}\right)+\mathbf{2 y}\left(\mathbf{f}-\mathbf{f}^{\prime}\right)+\mathbf{c}-\mathbf{c}^{\prime}=\mathbf{0} \\
& \text { or, } \quad \mathbf{S}-\mathbf{S}^{\prime}=\mathbf{0}
\end{aligned}
$

Length of Common Chord AB

$
\begin{aligned}
& \mathrm{AB}=2(\mathrm{AM}) \quad(\because \mathrm{M} \text { is the mid-point of } \mathrm{AB}) \\
& \quad=2 \sqrt{\left\{\left(\mathrm{C}_1 \mathrm{~A}\right)^2-\left(\mathrm{C}_1 \mathrm{M}\right)^2\right\}}
\end{aligned}
$

$\mathrm{C}_1 \mathrm{~A}=$ radius of the circle $\mathrm{S}=0$,
(i.e. $C_1 A=\sqrt{g^2+f^2-c}$ )
$\mathrm{C}_1 \mathrm{M}=$ length of perpendicular from $\mathrm{C}_1$ on common chord AB
The length of a common chord $A B$ of two circles is maximum when it is the diameter of the smaller circle between them.