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 \(x^2+y^2-6 x-2 y=0\)…
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
\(\left(\frac{33}{4}, \frac{20}{3}\right)\)
\(\left(\frac{33}{4}, \frac{10}{3}\right)\)
\(\left(\frac{33}{4}, \frac{-20}{3}\right)\)
\(\left(\frac{7}{6}, \frac{11}{6}\right)\)
Solution
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 \(\left(\frac{7}{6}, \frac{11}{6}\right)\)