Given:

$C_1\equiv x^2+y^2-4x-6y-12=0$

$C_2\equiv x^2+y^2+6x+18y+26=0$

Now, we know that length of tangent drawn from any point to circle is given by $\sqrt{S_1}$.

Now, it is given that, $\frac{\sqrt{S_1}}{\sqrt{S_2}}=\frac{2}{3}$

$\Rightarrow \frac{S_1}{S_2}=\frac{4}{9}$

$\Rightarrow 9S_1-4S_2=0$

$\Rightarrow 9\left(h^2+k^2-4h-6k-12\right)-4\left(h^2+k^2+6h+18k+26\right)=0$

$\Rightarrow 5h^2+5k^2-60h-126k-212=0$

$\Rightarrow h^2+k^2-12h-\frac{126}{5}k-\frac{212}{5}=0$

Therefore, locus is

$x^2+y^2-12x-\frac{126}{5}y-\frac{212}{5}=0$.