From the image below I know the vectors P1, P2 and P3. How can I find the point A which lies on the circle and line P2A which is a bisector (angle q and r are equal) of the lines P2P1 and P2B? Also how can I find the point B which is in the same direction as P2P3 and lies on the circle?

enter image description here

  • \$\begingroup\$ Down vote without a reason. How am suppose to know whats wrong with the question or let alone improve it? \$\endgroup\$ Apr 9, 2018 at 21:58
  • \$\begingroup\$ This is off topic. please consider using math.stackexchange.com instead for maths questions. \$\endgroup\$ Apr 10, 2018 at 10:57

4 Answers 4


First, let's name some more vectors so that the origin is at P2 :

$$v_1 = P_1 - P_2$$ $$v_3 = P_3 - P_2$$

We can find B by normalizing v3 and scaling it by the length of v1 (that is, the radius of the circle):

$$v_B = {\lVert v_1 \rVert \over \lVert v_3 \rVert} v_3$$ $$B = P_2 + v_B$$

Finally, we can sum both side vectors to get a bisector, and normalize and scale that to get A: $$v_A = {\lVert v_1 \rVert \over \lVert v_1 + v_B \rVert}(v_1 + v_B)$$ $$A = P_2 + v_A$$

You will probably find possible optimizations while implementing this.

  • \$\begingroup\$ Please what does || represent? I used the Unity tag, because I'm implementing this in a C# Script. \$\endgroup\$ Apr 9, 2018 at 21:57
  • \$\begingroup\$ @Containment it's the length of the vector inside :) \$\endgroup\$
    – Quentin
    Apr 10, 2018 at 8:12

P2P3.normalized*R = P2B;

Where R = P2P1.magnitude or R = P2A.magnitude; etc...

P2A = Quaternion.AngleAxis(Q, Vector3.forward)* P2P1 = Quaternion.AngleAxis(-r, Vector3.forward)* P2B;


Let \$\vec a\$ be \$P1- P2\$ and \$ \vec b\$ be \$P3-P2\$

Then point B is simply

$$ P2 +\frac{\vec b}{|\vec b|} \cdot radius$$

And point A is

$$ \vec c = (B + P1) / 2 - P2$$ $$ A = P2 + \frac{\vec c}{|\vec c|} \cdot radius$$

  • \$\begingroup\$ You're missing a factor, A lies on a unit circle around P2 here. \$\endgroup\$
    – Quentin
    Apr 9, 2018 at 15:35
  • \$\begingroup\$ @Quentin there you go \$\endgroup\$
    – Bálint
    Apr 9, 2018 at 15:39

Here's the answer strictly in unity's terms.

public Vector2 p1;
public Vector2 p2;
public Vector2 p3;
Vector2 b;
Vector2 a;

void Start() {
    float radius = (p1 - p2).magnitude;
    //Direction from p2 to p3
    Vector2 p2_p3_Dir = (p3 - p2).normalized;
    //Find point b
    b = p2 + p2_p3_Dir * radius;

    //Vector from b to p1
    Vector2 b_p1 = (p1 - b);
    //Distance from b to p1
    float b_p1_Distance = b_p1.magnitude;
    //Direction vector from b to p1
    Vector2 b_p1_Dir = b_p1.normalized;
    //Temporary point midway between b and p1
    Vector2 temp = b + b_p1_Dir * (b_p1_Distance/2);
    //Direction from p2 to temp
    Vector2 p2_temp_Dir = (temp - p2).normalized;
    //Finally, find a
    a = p2 + p2_temp_Dir * radius;


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .