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Home/ Questions/Q 294737
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T06:23:39+00:00 2026-05-12T06:23:39+00:00

I have a line from A to B and a circle positioned at C

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I have a line from A to B and a circle positioned at C with the radius R.

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What is a good algorithm to use to check whether the line intersects the circle? And at what coordinate along the circles edge it occurred?

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  1. Editorial Team
    Editorial Team
    2026-05-12T06:23:39+00:00Added an answer on May 12, 2026 at 6:23 am

    Taking

    1. E is the starting point of the ray,
    2. L is the end point of the ray,
    3. C is the center of sphere you’re testing against
    4. r is the radius of that sphere

    Compute:
    d = L – E ( Direction vector of ray, from start to end )
    f = E – C ( Vector from center sphere to ray start )

    Then the intersection is found by..
    Plugging:
    P = E + t * d
    This is a parametric equation:
    Px = Ex + tdx
    Py = Ey + tdy
    into
    (x – h)2 + (y – k)2 = r2
    (h,k) = center of circle.

    Note: We’ve simplified the problem to 2D here, the solution we get applies also in 3D

    to get:

    1. Expand
      x2 – 2xh + h2 + y2 – 2yk + k2 – r2 = 0
    2. Plug
      x = ex + tdx
      y = ey + tdy
      ( ex + tdx )2 – 2( ex + tdx )h + h2 +
      ( ey + tdy )2 – 2( ey + tdy )k + k2 – r2 = 0
    3. Explode
      ex2 + 2extdx + t2dx2 – 2exh – 2tdxh + h2 +
      ey2 + 2eytdy + t2dy2 – 2eyk – 2tdyk + k2 – r2 = 0
    4. Group
      t2( dx2 + dy2 ) +
      2t( exdx + eydy – dxh – dyk ) +
      ex2 + ey2 –
      2exh – 2eyk + h2 + k2 – r2 = 0
    5. Finally,
      t2( d · d ) + 2t( e · d – d · c ) + e · e – 2( e · c ) + c · c – r2 = 0
      Where d is the vector d and · is the dot product.
    6. And then,
      t2( d · d ) + 2t( d · ( e – c ) ) + ( e – c ) · ( e – c ) – r2 = 0
    7. Letting f = e – c
      t2( d · d ) + 2t( d · f ) + f · f – r2 = 0

    So we get:
    t2 * (d · d) + 2t*( f · d ) + ( f · f – r2 ) = 0

    So solving the quadratic equation:

    float a = d.Dot( d ) ;
    float b = 2*f.Dot( d ) ;
    float c = f.Dot( f ) - r*r ;
    
    float discriminant = b*b-4*a*c;
    if( discriminant < 0 )
    {
      // no intersection
    }
    else
    {
      // ray didn't totally miss sphere,
      // so there is a solution to
      // the equation.
      
      discriminant = sqrt( discriminant );
    
      // either solution may be on or off the ray so need to test both
      // t1 is always the smaller value, because BOTH discriminant and
      // a are nonnegative.
      float t1 = (-b - discriminant)/(2*a);
      float t2 = (-b + discriminant)/(2*a);
    
      // 3x HIT cases:
      //          -o->             --|-->  |            |  --|->
      // Impale(t1 hit,t2 hit), Poke(t1 hit,t2>1), ExitWound(t1<0, t2 hit), 
    
      // 3x MISS cases:
      //       ->  o                     o ->              | -> |
      // FallShort (t1>1,t2>1), Past (t1<0,t2<0), CompletelyInside(t1<0, t2>1)
      
      if( t1 >= 0 && t1 <= 1 )
      {
        // t1 is the intersection, and it's closer than t2
        // (since t1 uses -b - discriminant)
        // Impale, Poke
        return true ;
      }
    
      // here t1 didn't intersect so we are either started
      // inside the sphere or completely past it
      if( t2 >= 0 && t2 <= 1 )
      {
        // ExitWound
        return true ;
      }
      
      // no intn: FallShort, Past, CompletelyInside
      return false ;
    }
    
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