Discussion Forum

1)

The radical centre of the circle 

$x^{2}+y^{2}-4x-6y+5=0$, $  x^{2}+y^{2}-2x-4y-1=0$  and

$x^{2}+y^{2}-6x-2y=0=0 $ lies on the line 

Answer:

Option D

Explanation:

 $S_{1}= x^{2}+y^{2}-4x-6y+5=0$

$S_{2}= x^{2}+y^{2}-2x-4y-1=0$

$S_{3}= x^{2}+y^{2}-6x-2y=0$

 $S_{1}-S_{2}=0 \Rightarrow 2x+2y-6=0 \Rightarrow x+y-3=0 $ .....(i)

 $S_{2}-S_{3}=0 \Rightarrow 4x-2y-1=0$   ......(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 equation 18x-12y+1=0