Question

If the circles $x^2+y^2-2x-2y-7=0$ and $x^2+y^2+4x+2y+k=0$ cut orthogenally, then the length of the common chord of the circles is

• A. 2
• B. $12/\sqrt{13}$
• C. 8
• D. 5

Answer 1

Answer: (B)

Given, circles are
$S_1=x^2+y^2-2x-2y-7=0$
$S_2=x^2+y^2+4x+2y+k=0$
Here, $g_1=-1,f_1=-1.c_1=-7$
$g_2=2,f_2=1,c_2=k$

Equation of common chord is $S_1-S_2=0$
$\Rightarrow x^2+y^2-2x-2y-7-x^2-y^2-4x-2y-k=0$
$\Rightarrow -6x-4y-7—k=0$
$\Rightarrow 6x+4y+7+k=0\dots (i)$
Since, $S_1$ and $S_2$ cut orthogonally
$\therefore 2(g_1{g}_2+f_1{f}_2)=c_1+c_2$
$\Rightarrow 2(-2-1)=-7+k$
$\Rightarrow -6+7=k$
$\Rightarrow k=1$
Then, from Eqs. (i), we get
$6x+4y+8=0$
Now, length of the common chord
$r_1=\sqrt{1+1+7}=3$
$c_1=\left(1,1\right)$
Let $C_{1}M$ = perpendicular distance from centre $C_1(1,1)$ to the common chord
$6x+4y+8=0$
$\Rightarrow C_{1}M=\frac{|6+4+8|}{\sqrt{6^2+4^2}}=\frac{|18|}{5^2}=\frac{18}{2\sqrt{13}}=\frac{9}{\sqrt{13}}$
Now, $PQ=2PM=2\sqrt{{\left(C_{1}P\right)}^2-{\left(C_{1}M\right)}^2}$
$=2\sqrt{9-{\left(\frac{9}{\sqrt{13}}\right)}^2}$
$=2\sqrt{9-\frac{81}{13}}$
$=2\sqrt{\frac{36}{13}}=\frac{12}{\sqrt{13}}$