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
$S_{1} \equiv x^{2}+y^{2}-16 x+60=0 \,\,\,\,\dots(i)$
$S_{2} \equiv x^{2}+y^{2}-12 x+27=0 \,\,\,\,\,\dots(ii)$
and $S_{3} =x^{2}+y^{2}-12 y+8-0\,\,\,\,\,\dots(iii)$
The radical axis of circles (i) and (ii) is
$S_{1}-S_{2}=0$
$\Rightarrow \,\left(x^{2}+y^{2}-16 x+60\right)-\left(x^{2}+y^{2}-12 x+27\right)=0$
$\Rightarrow \,-4 x+33=0$
$\Rightarrow \, x=\frac{33}{4} \,\,\,\,\,\dots(iv)$
The radical axis of circles (ii) and (iii) is
$S_{2}-S_{3}=0$
$\Rightarrow \,\left(x^{2}+y^{2}-12 x+27\right)-\left(x^{2}+y^{2}-12 y+8\right)=0$
$\Rightarrow \, -12 x+12 y+19 \equiv 0\,\,\,\,\,\,\dots(v)$
On solving Eqs. (iv) and (v), we get radical centre $\left(\frac{33}{4}, \frac{20}{3}\right)$