Question

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

Answer 1

Give equation of circle-

$S:x^2+y^2+2x+2y+1=0$

$S_1:x^2+y^2+4x+3y+2=0$

Equation of common chord is-

$S-S_1=0$

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

$2x+y+1=0$

Thus the equation of common chord of the given circles is $2x+y+1=0$.

Now,

As we know that,

Length of common chord $=2\sqrt{r^2-d^2}$

Whereas $d$ and $r$ is the distance between the centre of any circle to the chord and radius of same circle respectively.

Therefore, Distance from centre of first circle to the chord-

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

Radius of first circle $\left(r\right)=1$

Therefore,

Length of chord $=2\sqrt{{\left(1\right)}^2-{\left(\frac{2}{\sqrt{5}}\right)}^2}=2\sqrt{\frac{5-4}{5}}=\frac{2}{\sqrt{5}}$

Hence the length of chord is $\frac{2}{\sqrt{5}}$.