geometry
I know the coordinates of line AB, I also have the coordinates of a point called C.
I need to find the coordinates of the start of a line that is perpendicular to AB and that would cross point C. (Point D)
Also the coordinates of point A are always (0, 0)
Suppose $A(a_1,a_2),B(b_1,b_2)$ and $D(d_1,d_2)$.
The line $(\epsilon_1)$ passing from these points has an equation of the form $Ax+By+C=0$.
Since $DC$ (lets call her $\epsilon_2$ ) is perpendicular to $\epsilon_1$ lets distinquish the following cases:
A. If $\epsilon_1$ has an equation of the form $x=x_0,x_0\in\mathbb{R}$. Then $\epsilon_2$ has an equation of the form $y=y_0,y_0\in\mathbb{R}$. You can easily find $y_0$ from $D$ and then find the coordinates of $C$ from the given length.
B. If $\epsilon_1$ has an equation of the form $y=y_0,y_0\in\mathbb{R}$. Then $\epsilon_2$ has an equation of the form $x=x_0,x_0\in\mathbb{R}$. You can easily find $x_0$ from $D$ and then find the coordinates of $C$ from the given length.
C. If $\epsilon_1$ isn't perpendicular to any of the axes $xx',yy'$ then $\epsilon_1$ has an equation of the form $y=ax+b$ (find $a,b$).Then, $\epsilon_2$ has an equation of the form $y-d_2=\lambda(x-d_1)$ where $\lambda\cdot a=-1$. The circle $C$ with center $D(d_1,d_2)$ and radius the given length $r$ has an equation of the form $(x-d_1)^2+(y-d_2)^2=r^2$. Solving the system of equations for $C$ and $\epsilon_2$ will give you the desired coordinates. (Notice: you'll find 2 points in each case!)
Assuming no typo in what you wrote and that I properly understood (this does not seem to be sure according to the comments I received after my initial answer), if you have point $A$ $(x_1,x_2)$ and point $B$ $(x_2,y_2)$ the square of the distance is given by $$d^2=(x_1-x_2)^2+(x_2-y_2)^2$$
I am sure that you can take from here.
Best Answer
Let denote $(x_M,y_M)$ the coordinates of a point $M$. The desired point $D$ is characterized by
$$\vec{MC}\cdot \vec{MA}=\vec{MC}\cdot \vec{MB}=0$$ which means using the coordinates: \begin{align}(x_C-x_M)(x_A-x_M)+(y_C-y_M)(y_A-y_M)&=0\\(x_C-x_M)(x_B-x_M)+(y_C-y_M)(y_B-y_M)&=0\end{align} so solve this system of two equations for the unknowns $x_M$ and $y_M$.