Question

Find the equation and the length of the common chord of the following circles:
$x^2+y^2+2x+2y+1=0;x^2+y^2+4x+3y+2=0$

Answer 1

The equation of the common chord is given by $S-S_1=0$, Where $S,S_1$ are two circles.

$x^2+y^2+2x+2y+1=0;x^2+y^2+4x+3y+2=0$ respectively.

$S-S_1=0$

$⟹(x^2+y^2+2x+2y+1)-(x^2+y^2+4x+3y+2)=0$

$-2x-y-1=0$

$i.e.,2x+y+1=0$ is the equation of the common chord.

Center of $S=0$ is $(-1,-1)$

Radius $=\sqrt{1+1-1}=1$

Length of the perpendicular from the center is given by,

$d=\left|\frac{ax+by+c}{\sqrt{a^2+b^2}}\right|$

$\therefore$ If $(-1,-1)$ is the center, then the length of the perpendicular to the chord is,

$d=\left|\frac{2(-1)+(-1)+1}{\sqrt{2^2+1^2}}\right|=\frac{2}{\sqrt{5}}$

Length of the Chord is $2\sqrt{r^2-d^2}$

$=2\sqrt{1-\frac{4}{5}}$

$=\frac{2}{\sqrt{5}}$