Find the radical centre of the circles ${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-7=0,2{\mathrm{x}}^2+2{\mathrm{y}}^2+3\mathrm{x}+5\mathrm{y}-9=0$ and ${\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{y}=0$.

Given circles are
$\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-7=0 ...\left(1\right)$
${\mathrm{S}}^{\mathrm{\prime }}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+\frac{3}{2}\mathrm{x}+\frac{5}{2}\mathrm{y}-\frac{9}{2}=0 ...\left(2\right)$
${\mathrm{S}}^{\prime \prime }\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{y}=0 ...\left(3\right)$
Radical axis of (1) & (2) is S – S’ = 0
$\begin{array}{l}\Rightarrow \left(4-\frac{3}{2}\right)x-\frac{5}{2}y+\left(\frac{9}{2}-7\right)=0\\ \Rightarrow \frac{5}{2}x-\frac{5}{2}y-\frac{5}{2}=0\Rightarrow x-y-1=0\dots \left(4\right)\end{array}$
Radical axis of (1) & (3) is S - S'' = 0
$\Rightarrow 4\mathrm{x}-\mathrm{y}-7=0 ...\left(5\right)$
Solving (4) & (5), we get
Radical centre of the circles = (2, 1)