Find the control points for a cubic bezier curve segment from point a to point c passing through point b.

findCubicControlPoints.ts

/**
 * Find the control points for a cubic segment from point a to point c passing through point b.
 * @param a The curve's start point
 * @param b The point to curve through
 * @param c The curve's end point
 */
export function findCubicControlPoints(
  a: { x: number; y: number },
  b: { x: number; y: number },
  c: { x: number; y: number }
) {
  // center of the circle ABC
  const ctr = {
    x:
      ((a.x * a.x + a.y * a.y) * (c.y - b.y) +
        (b.x * b.x + b.y * b.y) * (a.y - c.y) +
        (c.x * c.x + c.y * c.y) * (b.y - a.y)) /
      ((a.x * (b.y - c.y) - a.y * (b.x - c.x) + b.x * c.y - c.x * b.y) * -2),
    y:
      ((a.x * a.x + a.y * a.y) * (b.x - c.x) +
        (b.x * b.x + b.y * b.y) * (c.x - a.x) +
        (c.x * c.x + c.y * c.y) * (a.x - b.x)) /
      ((a.x * (b.y - c.y) - a.y * (b.x - c.x) + b.x * c.y - c.x * b.y) * -2)
  };

  // tangent line through b
  const tan1 = { x: b.x - (b.y - ctr.y), y: b.y + (b.x - ctr.x) };
  const tan2 = { x: b.x + (b.y - ctr.y), y: b.y - (b.x - ctr.x) };

  // normalized slope of tangent line
  const tlength = Math.hypot(tan1.y - tan2.y, tan1.x - tan2.x);
  const dx = (tan2.x - tan1.x) / tlength;
  const dy = (tan2.y - tan1.y) / tlength;

  // find t
  const dAB = Math.hypot(a.y - b.y, a.x - b.x);
  const dBC = Math.hypot(c.y - b.y, c.x - b.x);
  const dAC = Math.hypot(a.y - c.y, a.x - c.x);
  const t = dAB / (dAB + dBC);

  // points e1 and e2 on tangent line from circle (through b)
  const aCA = Math.atan2(c.y - a.y, c.x - a.x);
  const aBA = Math.atan2(b.y - a.y, b.x - a.x);
  const ang = aCA - aBA;

  const bc = ((ang < 0 || ang > Math.PI ? -1 : 1) * dAC) / 3;
  const de1 = t * bc;
  const de2 = (1 - t) * bc;

  const e1 = { x: b.x + de1 * dx, y: b.y + de1 * dy };
  const e2 = { x: b.x - de2 * dx, y: b.y - de2 * dy };

  // point d
  const t1 = Math.pow(1 - t, 3);
  const b1 = Math.pow(t, 3) + t1;
  const u = t1 / b1;
  const s = Math.abs((b1 - 1) / b1);

  const d = {
    x: b.x + (b.x - (u * a.x + (1 - u) * c.x)) / s,
    y: b.y + (b.y - (u * a.y + (1 - u) * c.y)) / s
  };

  // interpolated point on line D/e1
  const v1 = {
    x: d.x + (e1.x - d.x) / (1 - t),
    y: d.y + (e1.y - d.y) / (1 - t)
  };

  // interpolated point on line D/e2
  const v2 = {
    x: d.x + (e2.x - d.x) / t,
    y: d.y + (e2.y - d.y) / t
  };

  // control point1, interpolated point on line a/v1
  const c1 = {
    x: a.x + (v1.x - a.x) / t,
    y: a.y + (v1.y - a.y) / t
  };

  // control point2, interpolated point on line c/v2
  const c2 = {
    x: c.x + (v2.x - c.x) / (1 - t),
    y: c.y + (v2.y - c.y) / (1 - t)
  };

  return { a, c1, c2, b };
}

Demo at https://codesandbox.io/s/find-handle-for-cubic-curve-single-fn-nirsnv