Question

The length of the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ is

• A. $\frac{9}{2}$
• B. $2\sqrt{2}$
• C. $3\sqrt{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

Let $S_1\equiv x^2+y^2+2x+3y+1=0$
and $S_2\equiv x^2+y^2+4x+3y+2=0$,
then equation of common chord is $S_2-S_1=0$

$\Rightarrow 2x+1=0$
Here, $C_1\left(-1,-\frac{3}{2}\right),r_1=\frac{3}{2}=C_{1}P$
and $C_2\left(-2,-\frac{3}{2}\right),r_2=\frac{\sqrt{17}}{2}$
$C_{1}M=$ Perpendicular distance from $C_1$ to the common chord $2x+1=0$
$\Rightarrow C_{1}M=\frac{|-2+1|}{\sqrt{2^2}}=\frac{1}{2}$
Now, $PQ=2PM=2\sqrt{{\left(C_{1}P\right)}^2-{\left(C_{1}M\right)}^2}$
$=2\sqrt{{\left(\frac{3}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}=2\sqrt{\frac{9}{4}-\frac{1}{4}}$
$=2\sqrt{2}$