Question

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

• A. $(13,33 / 4)$
• B. $(33 / 4,-13)$
• C. $(33 / 4,13)$
• D. None of these

Answer 1

Answer: (D)

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