Question

The radical axis of the circles $x^2+y^2+4x-6y=12$ and $x^2+y^2+2x-2y-1=0$ divides the line segement joining the centres of the circles in the ratio

• A. $27:17$
• B. $3:7$
• C. $-27:17$
• D. $-3:7$

Answer 1

Answer: (C)

$-27:17$

As we know, radical axis divides the line segment joining the centres of the circles in ratio of there radii.
So, $x^2+y^2+4x-6y=12$ and $x^2+y^2+2x-2y-1=0$
Radius $r_1=5$ and radius $r_2=\sqrt{3}$
$c_1=(-2,3),c_2=(-1,1)$
Equation of radical axis is $S_1-S_2$
i.e. $2x-4y-11=0$ ...... $\left(1\right)$
Equation of line passing through center of the circle
$y-1=-2(x+1)$
$y+2x+1=0$ ...... $\left(2\right)$
From (1) & (2)

$5y+12=0$
$y=\frac{-12}{5},x=\frac{7}{10}$
So, $c_1{c}_2=0$

$(\frac{7}{10},\frac{-12}{5})$ is the radical axis
Let $O$ divides $c_1{c}_2$ in $1:k$ ratio then
$\frac{7}{10}=\frac{-1-k(-2)}{1-k}$
$7-7k=20k-10\Rightarrow k=\frac{17}{27}$ externally
Since, it cuts externally

$\Rightarrow 1:k=-27:17$