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
$<br/>\begin{array}{l}<br/>S_1\equiv x^2+y^2-16x+60=0\dots \text{ (i) }\\ <br/>S_2\equiv x^2+y^2-12x+27=0\dots \text{ (ii) And }\\ <br/>S_3\equiv x^2+y^2-12y+8=0\dots \text{ (iii) The radical axis }<br/>\end{array}<br/>$
$S_3=x^2+y^2-12y+8=0$... (iii) The radical axis of circles (i) and (ii) is $S_1-S_2=0\Rightarrow$
$\left(x^2+y^2-16x+60\right)-\left(x^2+y^2-12x+27\right)=0$ $\Rightarrow -4x+33=0\Rightarrow x=\frac{33}{4}..$ (iv)
The radical axis of circles (ii) and (iii) is
$S_2-S_3=0\Rightarrow$ $\left(x^2+y^2-12x+27\right)-\left(x^2+y^2-12y+8\right)=0\Rightarrow$ $-12x+12y+19=0$ ? (v) On solving Eqs. (iv) and (v), wef get radical centre
$\left(\frac{33}{4},\frac{20}{3}\right)$.