Find arc center from tangent lines and 'rounding value' Simple and common question: I want to round two intersecting lines with arc, so I need to know its center point.

I have defined:

*

*AP - first line

*BP - second line

*|PR| - rounding scalar value, so the arc stars on R point

How to find C - center point of arc?
 A: If you gave vectors, notice that $\vec{CR} \cdot \vec{PB} = \vec{CQ} \cdot \vec{PA} = 0$ where $Q = P + \frac{|PR|}{|PB|}\vec{PA}$.
A: This very look likes a computation useful to code an arcTo implementation. I'm adding my version in modern C# syntax, assuming $(x0,y0)$ to the coordinates of the point $a$, $(x1,y1)$ to the coordinates of the point $P$ and $(x2,y2)$ to the coordinates of the point $B$:
static (double xc, double yc) ComputeArcCenter(double x0, double y0, double x1, double y1, double x2, double y2, double r)
{
    // Reference https://math.stackexchange.com/questions/191942/find-arc-center-from-tangent-lines-and-rounding-value

    double x1_0 = x0 - x1;
    double y1_0 = y0 - y1;
    double x1_2 = x2 - x1;
    double y1_2 = y2 - y1;

    // Compute the tagent points
    double norm1 = Math.Sqrt(x1_0 * x1_0 + y1_0 * y1_0);
    double norm2 = Math.Sqrt(x1_2 * x1_2 + y1_2 * y1_2);

    double x1t = x1 + x1_0 / norm1 * r;
    double y1t = y1 + y1_0 / norm1 * r;
    double x2t = x1 + x1_2 / norm2 * r;
    double y2t = y1 + y1_2 / norm2 * r;

    // Compute a two-point form -(y2–y1)*(x-x1) + (x2-x1)*(y-y1) = 0 and
    // then find the equation of perpendicular on the point (x1,y1) with b*x - a*y + a * y1 − b * x1 = 0

    // Compute the coefficientes of a line passing through (x1t, y1t) and perpendicular to the arc tangent
    double a0t = x1_0;
    double b0t = y1_0;
    double c0t = - x1_0 * x1t - y1_0 * y1t;

    // Compute the coefficientes of a line passing through (x2t, y2t) and perpendicular to the arc tangent
    double a2t = x1_2;
    double b2t = y1_2;
    double c2t = - x1_2 * x2t - y1_2 * y2t;

    // https://www.cuemath.com/geometry/intersection-of-two-lines/
    double xc = (b0t * c2t - b2t * c0t) / (a0t * b2t - a2t * b0t);
    double yc = (c0t * a2t - c2t * a0t) / (a0t * b2t - a2t * b0t);

    return (xc, yc);
}

One can then plug a method to compute the control points of a Bézier curve to draw the arc.
