Given circles are

$S_1\equiv x^2+y^2-8x+2y+8=0 ...\left(1\right)$

$S_2\equiv x^2+y^2+6x+8y-24=0 ...\left(2\right)$

& $S_3\equiv x^2+y^2-2x+2y+2=0 ...\left(3\right)$

$\therefore$Radical axis of $S_1 \& S_2$is given by $S_1-S_2=0$

$\Rightarrow -14x-6y+32=0$

$\Rightarrow 7x+3y-16=0 ...\left(4\right)$

Radical axis of $S_{2 }\& S_3$ is given by $S_2-S_3=0$

$\Rightarrow 8x+6y-26=0$

$\Rightarrow 4x+3y-13=0 ...\left(5\right)$

As, common point of intersection of the radical axis of three circles taken two at a time is radical centre of three circles.So here radical centre is point of intersection of $\left(4\right) \& \left(5\right)$.

Equation $\left(4\right)$$-$equation $\left(5\right)$, we have

$\left(7x+3y-16\right)-\left(4x+3y-13\right)=0$

$3x-3=0$

$\therefore x=1$

Substituting $x=1$in $\left(5\right)$, we get

$4+3y-13=0$

$\therefore y=3$

$\therefore a=1, b=3$

$\therefore a+b=1+3=4$