Draw a picture. Let our triangle be $ABC$, with right angle at $C$. Let the legs be $a$ and $b$. Let the median to side $a$ have length $13$. By the Pythagorean Theorem, we have $\frac{a^2}{4}+b^2=13^2$. Similarly, $a^2+\frac{b^2}{4}=19^2$. Solve.
Or better, don't solve. Add. We get $\frac{5}{4}(a^2+b^2)=13^2+19^2$.
To solve this problem we need to draw two auxiliary lines, i.e. the angle bisectors of $\measuredangle A$ and $\measuredangle C$ as shown in the diagram. Both these lines passes through the incenter $I$ of the triangle $ABC$. By the way, $Q$ is the intersection point of side $BC$ and line $MN$, which is parallel to the side $CA$. Consequently, we can state that the two triangles $API$ and $CIQ$ are isosceles triangles.
It is given $AB=30$, $MP=8$, and $PN=25$. Let $AP=x$, $CA=3y$, and $CQ=z$.
You have already found one of the equations we need to solve the problem, i.e.
$$30x-x^2=200 \tag{1}$$
Equation (1) has two roots, i.e. $x=10$ and $x=20$. First, we take the former and check whether it leads us to a solution. So, from now on, we have $AP=x=10$, which makes $PB=20$. As consequence, we have $AP:PB=1:2$. Since $PQ$ is parallel to $CA$, $CQ:QB=1:2$ as well.
Since $API$ is an isosceles triangle, we have $IP=AP=10$. In a similar vein, since $CIQ$ is an isosceles triangle, we can deduce $IQ=CQ=z$. Since the two lines $AC$ and $PQ$ are parallel to each other and $AP:PB=1:2$, we can write
$PQ=2y$. Therefore, we are able to form the following equation.
$$PQ=PI+IQ \qquad\rightarrow\qquad 2y=10+z \tag{2}$$
Now, intersecting chord theorem can be used to obtain the following equation.
$$MQ\cdot QN=\left(MP+PQ\right)\left(PN-PQ\right)=CQ\cdot QB \quad\rightarrow\quad \left(8+2y\right)\left(25-2y\right)=z\cdot 2z $$
$$200+34y-4y^2=2z^2 \tag{3}$$
When we substitute the value of $z$ from the equation (2) in equation (3), we get,
$$200+34y-4y^2=2\left(2y-10\right)^2=8y^2+200-80y \qquad\rightarrow\qquad 12y^2-114y=0 \tag{4}$$
Equation (4) has two roots, i.e. $y=0$ and $y=9.5$, where the latter is the only acceptable solution. When we insert this value of $y$ into the equation (2), we get $z=9$.
Therefore, the the length of the two sides $CA$ and $BC$ can be written as
$$CA=3y=3\times 9.5=28.5\quad \mathrm{and}$$
$$BC=3z=3\times 9=27.$$
Now, it is up to you to work out the case where $x=20$.
Best Answer
The incenter has trilinear coordinates $1:1:1$, which you can convert to barycentric coordinates $a:b:c$, where $a=\lVert B-C\rVert, b=\lVert C-A\rVert, c=\lVert A-B\rVert$ are the edge lengths of the triangle. The incenter therefore has coordinates
$$\frac{a}{a+b+c}A + \frac{b}{a+b+c}B + \frac{c}{a+b+c}C$$
no matter the dimension of the space where these corners $A,B,C$ live.
That won't work. As Blue already pointed out in a comment, the incircle center won't lie on that median. If you want to start with the computed incircle radius, you could take a line parallel to the edge but shifted by the given distance into the direction of the other edge, and do so for two edges. But the above is much simpler.
The centroid does lie on the medians, but for that you'd simply intersect two medians and you don't get a reasonable circle associated with that triangle center.