How to find the vertices with integral coordinates of a triangle knowing that center and inradius? I want to find the length of the sides of a triangle (not right triangle) with center $C(1; 2)$ and inradius $R=5$. How can I do?
I tried.


*

*Write the equation of the circle with center $C(1; 2)$ and inradius $R=5$;

*Choose three points $A(-3,-1)$, $B(4,6), $C(5,5) lies on circle;

*Write the equation of the tangents at the point $A$, $B$, $C$;

*This tangents cut at three points $M$, $N$, $P$. 
A problem is opened, how to find the vertices with integral coordinates?

 A: To make this question easier, I'll put the center of the incircle in the origin. Any integral solution can be easily transferred to the location you indicated. Let $M,N,P$ be the corners of the triangle, as used in your question. Let the coordinates of $M$ be $(M_x,M_y)\in\mathbb Z^2$. One important quantity in your scenario is
$$d:=M_x^2+M_y^2-25$$
As you can see, as long as $M$ is outside the circle, that quantity will be positive. Its square root can be used to describe the tangents from $M$ to the circle:
\begin{align*}
t_1 &= \left\{(x,y)\;\middle\vert\;
(5M_x+\sqrt dM_y)x + (5M_y-\sqrt dM_x)y = 5\left(M_x^2+M_y^2\right) \right\} \\
t_2 &= \left\{(x,y)\;\middle\vert\;
(5M_x-\sqrt dM_y)x + (5M_y+\sqrt dM_x)y = 5\left(M_x^2+M_y^2\right) \right\}
\end{align*}
The way I obtained these tangents is by performing the dual of a conic-line intersection, as described in section 11.3 of Perspectives on Projective Geometry by Richter-Gebert.
If these tangents are to cross any points with integer coordinates, their slopes have to be rational (or $\infty$ for vertical lines). These slopes are
$$-\frac{5M_x\pm\sqrt dM_y}{5M_y\mp\sqrt dM_x}$$
There are two conceivable ways these slopes might be rational: either because $\sqrt d$ is rational, or because numerator and denominator are rational multiples of the same irrational number. If one treats $\mathbb Q[\sqrt d]$ as a two-dimensional $\mathbb Q$-vectorspace, then this translates into the requirement that the two vectors have to be linearily dependent, which here means
$$\det\begin{pmatrix}
5M_x & \pm M_y \\
5M_y & \mp M_x
\end{pmatrix}=\mp5\left(M_x^2+M_y^2\right)=0$$
So the only other way would mean $M$ is the origin, which is of course impossible. So we are left with the requirement that $\sqrt d$ be rational, so $d$ must be a square number.
Now one possible way to tackle this problem is iterating over all integral coordinates in a given range, and checking whether the resulting $d$ is square. One can build up a collection of possible corner points, and using the coordinates of the tangents, one can see which corners could belong to the same triangle. The $t_1$ of one point must match the $t_2$ of the other edge, but scalar multiples of the coefficients describe the same line. I've implemented such an enumeration. Many results contain $5$ as one of the coordinates of one of the points. But not all do, so one of the more interesting results is
\begin{align*}
M&=(-11, -2) & N&=(  1,  7) & P&=( 13, -9)
\end{align*}
After moving $C$ back to where you defined it, this becomes
\begin{align*}
M&=(-10, 0) & N&=(  2,  9) & P&=( 14, -7)
\end{align*}

However, the solution above has aright angle at $N$, so as your question asks for a slution without a right angle, this one is not acceptable. Thanks to @coffeemath for making me aware of this condition. It seems that at least if the absolute value of the coordinates is restricted to no more than $1000$, then all solution without a right angle neccessarily have either a horizontal or a vertical tangent. So here is another go, again first in the untranslated situation:
\begin{align*}
M&=(-11, -10) & N&=( -3,  5) & P&=( 25, 5)
\end{align*}
After moving $C$ back to where you defined it, this becomes
\begin{align*}
M&=(-10, -8) & N&=(  -2,  7) & P&=( 26, 7)
\end{align*}

I also adapted my program to filter out the right-angled triangles automatically. There are still many possible solutions remaining, but none with all coordinates less than $25$ in absolute value, in the coordinate system where the circle center is the origin.
