Question

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

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

Answer 1

Answer: (D)

$18x-12y+1=0$

Give equations $x^2+y^2-4x-6y+5=0,x^2+y^2-2x-4y-1=0,x^2+y^2-6x-2y=0$

R.A of (a) and (b) is $-2x-2y+6=0$ $\Rightarrow$ $x+y=3$
R.A of (a) and (c) is $2x-4y+5=0$ $\Rightarrow x=3-\frac{11}{6}=\frac{7}{6}$
Thus $-6y=-11\Rightarrow y=\frac{11}{6}$

Radical centre $(\frac{7}{6},\frac{11}{6})$

From the above options, it can be seen that the above point lies on the line $18x-12y+1=0$ as the LHS $=$ RHS in this case.