Coordinates of the point on the circle inscribed in a square I try to find a way to calculate coordinates of a point nested on a circle inscribed in a square. The available variables, are:
1) side length of the square = 100;
2) circle radius = 50;
3) angle (a) = 45 degrees, but it can vary (e.g. 21, 15.3, etc.)
What I'm looking for in the situation described above are X2 x and y coordinates on the given square.

I'm out of ideas, so I'll be appreciate for any help.
 A: put your square in a cartesian coordinate system an compute the equation of the incircle in the form
$(x-x_M)^2+(y-y_M)^2=R^2$
$M(x_M,y_M)$ is the middlepoint of the incircle with the coordinates   $x_M,y_M$
A: If we choose as origin the bottom-left vertix, and call the radius with the letter $\ r$:
$\ X_{2}(c_{x},c_{y})\iff X_{2}(50+c,r-c)$, where $c=r \sin(45°)=\frac{r}{\sqrt2}$
A: (This may be using a sledgehammer to crack a nut...)
$X_2$ can be interpreted as a clockwise rotation of $X_1$ around $X_0$ with an angle of $a$.
This can be expressed as a rotation matrix:
$$
R(a)=
\begin{bmatrix}
+\cos a & -\sin a \\
+\sin a & +\cos a \\
\end{bmatrix}
$$
If you multiply this matrix with a column vector, then the result is a clockwise (*) rotation of the vector by an angle $a$ around the origin $(0,0)$:
$$
\begin{bmatrix}
+\cos a & -\sin a \\
+\sin a & +\cos a \\
\end{bmatrix}\begin{bmatrix}
x \\
y \\
\end{bmatrix} = (x \cos a - y \sin a, x \sin a + y \cos a)
$$
But we do not want to rotate around $(0,0)$ but $X_0$, so we need a coordinate transformation where $X_0$ corresponds to $(0,0)$: 
$$\hat X_1 = X_1-X_0 = (100,50) - (50,50) = (50,0)$$
Now we rotate $\hat X_1$ using the matrix above: 
$$\begin{align}
\hat X_2 & = (\hat X_{1x} \cos 45° - \hat X_{1y} \sin 45°, \hat X_{1x} \sin 45° + \hat X_{1y} \cos 45°) \\
& = (50 \cos 45° - 0 \sin 45°, 50 \sin 45° + 0 \cos 45°) \\
& = (50 \cos 45°, 50 \sin 45°) \\
& \approx (35.36, 35.36)
\end{align}$$
And reverse the coordinate transformation: $$X_2 = \hat X_2 + X_0 \approx (85.36, 85.36)$$
By the way: Using homogenous coordinates, we could even combine all of this into a single transformation matrix.
(*) In this example, it's a clockwise rotation because the coordinate system is left-handed. In a right-handed system, it would be a counterclockwise rotation.
A: Hope the following illustration helps:-
A: Let the perpendicular from $X_2$ to $X_0X_1$ be $X$. Then $\cos a=X_0X/50$ and $\sin a=X_2X/50$. From the way you have defined the coordinates so that moving right increases the horizontal coordinate and moving down increases the vertical coordinate, the coordinates of $X_2$ are $$(50+50\cos a,50+50\sin a).$$
