Equation of the given circles can be written as $C_1:{\left(x-1\right)}^2+{\left(y-3\right)}^2=1$  having centre at $\left(1,3\right)$  and radius 1 and $C_2:{\left(x+3\right)}^2+{\left(y-1\right)}^2=9$  having centre at $\left(-3,1\right)$  and radius 3.

Both the circles clearly touch $x=0andy=4$

Now,                  $\left|\frac{3\times 1+4\times 3-10}{\sqrt{3^2+4^2}}\right|=1$

And                    $\left|\frac{3\left(-3\right)+4\left(1\right)-10}{\sqrt{3^2+4^2}}\right|=3$

So, both the circles touch $3x+4y=10$  also.