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