Question

The radical centre of the circles $x^{2}+y^{2}-4\,x-6 \,y+5=0$ $x^{2}+y^{2}-2\,x-4 \,y-1=0$ and $x^{2}+y^{2}-6 \,x-2 \,y=0=0$ lies on the line

• A. x + y - 5 = 0
• B. 2 x - 4 y + 7 = 0
• C. 4 x - 6 y + 5 = 0
• D. 18 x - 12 y + 1 = 0

Answer 1

Answer: (D)

18 x - 12 y + 1 = 0

Answer 2

We have,
$S_{1}=x^{2}+y^{2}-4 \,x-6 \,y+5=0$
$S_{2}=x^{2}+y^{2}-2 \,x-4 \,y-1=0$
$S_{3}=x^{2}+y^{2}-6 \,x-2\,y=0$
$S_{1}-S_{2}=0 \Rightarrow 2 \,x+2 \,y-6=0 \Rightarrow x+y-3=0 \ldots (i)$
$S_{2}-S_{3}=0 \Rightarrow 4 \,x-2 \,y-1=0 \ldots (ii)$
Solving Eqs. (i) and (ii), we get
$x=\frac{7}{6}, y=\frac{11}{6}$
$\left(\frac{7}{6}, \frac{11}{6}\right)$ satisfies the equations $18\, x-12\, y+1=0$