Given:

$$\begin{gathered} S_1\equiv x^2+y^2-4=0 \\ {S}_2\equiv x^2+y^2-2x+3y=0 \\ {S}_3\equiv x^2+y^2+7y-18=0 \end{gathered}$$

We know if the length of tangents drawn from an external points to three circles are equal, then that point is the radical centre.

First we will find the radical axes which is 

$S_1-S_2=0 \text{or} S_2-S_3=0$

We take,

$S_1-S_2=0$

$\Rightarrow 2x-3y-4=0 ...\left(i\right)$

And, $S_1-S_3=0$

$-7y+18-4=0 ...\left(ii\right)$

Solving $\left(i\right) \& \left(ii\right)$, we get

$x=5 \& y=2$.