The radical centre of the circles $x^2+y^2-4 x-6 y+5=0$, $x^2+y^2-2 x-4 y-1=0$ and $x^2+y^2-6 x-2 y=0$ is equal to

Question

The radical centre of the circles $x^2+y^2-4 x-6 y+5=0$, $x^2+y^2-2 x-4 y-1=0$ and $x^2+y^2-6 x-2 y=0$ is equal to, $\left(\frac{33}{4}, \frac{20}{3}\right)$, $\left(\frac{33}{4}, \frac{10}{3}\right)$, $\left(\frac{33}{4}, \frac{-20}{3}\right)$, $\left(\frac{7}{6}, \frac{11}{6}\right)$

MARKS App · Accepted Answer

Equation of given circles $\begin{aligned} & S_1: x^2+y^2-4 x-6 y+5=0 \ & S_2: x^2+y^2-2 x-4 y-1=0 \end{aligned}$ and $\quad S_3: x^2+y^2-6 x-2 y=0$ $\therefore$ Radical axis of circles $S_1$ and $S_2$ is $2 x+2 y-6=0 \Rightarrow x+y=3$...(i) Similarly, the radical axis of circles $S_2$ and $S_3$ is $-4 x+2 y+1=0$...(ii) $\because$ Radical centre is point of concurrency of radical axes, from radical axes Eqs. (i) and (ii), we get $y=\frac{11}{6} \text { and } x=\frac{7}{6}$ $\therefore$ Radical centre is $\left(\frac{7}{6}, \frac{11}{6}\right)$

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