The radical centre of the three circles $x^2+y^2-1=0, x^2+y^2-8 x+15=0$ and $x^2+y^2+10 y+24=0$ is
The radical centre of the three circles $x^2+y^2-1=0, x^2+y^2-8 x+15=0$ and $x^2+y^2+10 y+24=0$ is
$\left(2, \frac{-5}{2}\right)$
$\left(2, \frac{5}{2}\right)$
$\left(-2, \frac{5}{2}\right)$
$\left(-2, \frac{-5}{2}\right)$
Solution
Let $S_1: x^2+y^2-1=0$ ...(i)
$S_2: x^2+y^2-8 x+15=0$ ...(ii)
and $S_3: x^2+y^2+10 y+24=0$ ...(iii)
From Eqs. (i) and (ii),
$8 x-15=1$
$\Rightarrow \quad x=2$
Now, from Eqs. (i) and (iii),
$-10 y-24=1$
$\Rightarrow-10 y=25 \Rightarrow y=\frac{-5}{2}$
Hence, the radical centre is $\left(2, \frac{-5}{2}\right)$.