Question

The radical centre of the circles
$x^2+y^2-16x+60=0$ ,
$x^2+y^2-12x+27=0$
and $x^2+y^2-12y+8=0$ is

• A. $\left(13,\frac{33}{4}\right)$
• B. $\left(\frac{33}{4},-13\right)$
• C. $\left(\frac{33}{4},13\right)$
• D. None of these

Answer 1

Answer: (D)

Given, equation of the circles are
$S_1\equiv x^2+y^2-16x+60=0\dots (i)$
$S_2\equiv x^2+y^2-12x+27=0\dots (ii)$
and $S_3=x^2+y^2-12y+8-0\dots (iii)$
The radical axis of circles (i) and (ii) is
$S_1-S_2=0$
$\Rightarrow \left(x^2+y^2-16x+60\right)-\left(x^2+y^2-12x+27\right)=0$
$\Rightarrow -4x+33=0$
$\Rightarrow x=\frac{33}{4}\dots (iv)$
The radical axis of circles (ii) and (iii) is
$S_2-S_3=0$
$\Rightarrow \left(x^2+y^2-12x+27\right)-\left(x^2+y^2-12y+8\right)=0$
$\Rightarrow -12x+12y+19\equiv 0\dots (v)$
On solving Eqs. (iv) and (v), we get radical centre $\left(\frac{33}{4},\frac{20}{3}\right)$