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

(d)
The radical axis of circle Ist and IInd is
$S_{1} - S_{2} = 0$
$\Rightarrow \ \ \ \left( x^{2} + y^{2} - 16x + 60 \right) - \left( x^{2} + y^{2} - 12x + 27 \right) = 0$
$\Rightarrow \ \ - 4x + 33 = 0\ \ \ \Rightarrow \ \ \ x = \frac{33}{4}$ …(i)
The radical axis of circle IInd and IIIrd is
$S_{2} - S_{3} = 0$
$\Rightarrow \ \ \ \left( x^{2} + y^{2} - 12x + 27 \right) - \left( x^{2} + y^{2} - 12y + 8 \right) = 0$
$\Rightarrow \ \ - 12x + 12y + 19 = 0$ …(ii)
$\therefore$ From Eqs. (i) and (ii), we get radical centre $\left( \frac{33}{4},\frac{20}{3} \right)$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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