Hello Ernesto,
Let's orient a circle or radius $r$ centered at the origin, and place a point on the $x$-axis outside of the circle, and let it's distance from the origin be $d$. Drawing a tangent line from this point to the circle (in the first quadrant), we form a right triangle, since the tangent line is perpendicular to the radius of the circle drawn to the tangent point. Let $\ell$ be the length of this line segment. Hence, we may state:
$$r^2+\ell^2=d^2$$
Solving for $\ell$, we obtain:
$$\ell=\sqrt{d^2-r^2}$$
Now, observing that there is also a fourth quadrant tangent point, we may state that the sum $S$ of the lengths of the tangent line segments is:
$$S=2\sqrt{d^2-r^2}$$
Now, if we do this for two circles, one of radius $r_1$ and the other of radius $r_2$, we then have:
$$S=2\left(\sqrt{d^2-r_1^2}+\sqrt{d^2-r_2^2} \right)$$
In order to ease solving for $d$, let's write the equation as:
$$S-2\sqrt{d^2-r_2^2}=2\sqrt{d^2-r_1^2}$$
Squaring both sides of the equation, there results:
$$S^2-4S\sqrt{d^2-r_2^2}+4\left(d^2-r_2^2 \right)=4\left(d^2-r_1^2 \right)$$
Distributing and combining like terms, and isolating the term with the remaining radical we obtain:
$$4S\sqrt{d^2-r_2^2}=4r_1^2-4r_2^2+S^2$$
Squaring again, we obtain:
$$16S^2\left(d^2-r_2^2 \right)=\left(4r_1^2-4r_2^2+S^2 \right)^2$$
$$d^2=\frac{\left(4r_1^2-4r_2^2+S^2 \right)^2}{16S^2}+r_2^2=\frac{\left(4\left(r_1^2-r_2^2 \right)+S^2 \right)^2+\left(4r_2S \right)^2}{(4S)^2}$$
Now, since this will hold for any point that is a distance $d$ from the origin, which describes a circle centered at the origin having radius $d$, we then find the locus of points is:
$$x^2+y^2=\frac{\left(4\left(r_1^2-r_2^2 \right)+S^2 \right)^2+\left(4r_2S \right)^2}{(4S)^2}$$
Now, using the data given with the problem:
$$r_1=2,\,r_2=3,\,S=5$$
We find the locus of points is:
$$x^2+y^2=\left(\frac{\sqrt{145}}{4} \right)^2$$