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Finding the Intersection of Two Circles

Date: 08/19/99 at 22:44:21
From: Tom Groff
Subject: Cad Geometry

I have been writing a simple 2D wireframe CAD application in Visual 
Basic. I have resolved most of the intersection problems (Lines, Arcs 
and Circles) that I have come across even though my math is poor. I 
have stubbed my mental toe on the intersection points of two circles. 
My function checks for the four possible situations and deals with 
them in the following Pseudo Code.

   If Circ1 center point = Circ2 center point
         Intersection = false          'Circles are concentric.
      elseif Distance between centers = Circ1 radius + Circ2 radius
            Intersection = true        'Circles are tangent.
            x = (cx1+cx2)/2            'Intersection is the midpoint
            y = (cy1+cy2)/2            'between the circle center pts.
         elseif Distance between centers < Circ1 radius + Circ2 radius
               Intersection = false    'Circles do not meet.
               Intersection = true     'Circles definitely intersect.

Since we know the distance between the circle centers and the two 
circle radii, we have a non-right triangle. The apex of this triangle 
is one of the intersections of the two circles. Using the ArcCos() 
function, I can find the angle of the first radius triangle leg to the 
circle center point.

      Ang = Arccos(((DistC^2)+(Rad1^2)-(Rad2^2))/(2*DistC*Rad1)) 

Using more trig I can determine the distance to the midpoint between 
the circles.

      CenDist = Cos(Ang) * Rad1
      Dist    = Sin(Ang) * Rad1

Finally I walk up the line between the center points (CenDist).

   DistRatio = -(CenDist / hypot(Abs(cx1 - cx2), Abs(cy1 - cy2)))
   x1 = ((cx1 - cx2) * (DistRatio)) + cx1
   y1 = ((cy1 - cy2) * (DistRatio)) + cy1

This gives me the point on the CenterLine in line with the two 
intersection points. Logic says I should be able to find the inverse 
slope of the CenterLine, drive it through the new point on the 
CenterLine and walk out "Dist" to my circle intersection points.

   Sab = (cy1 - cy2) / (cx1 - cx2)  'Slope of the existing line
   Iab = cy1 - (Sab * cx1)          'Y-int. of the existing line

   Scd = -(1 / Sab)                 'Inv. slope of the existing line
   Icd = y1 - (Scd * x1)            'Y-int. of calculated line

What is frustrating is that after all this work I found I do not know 
how to define a new point at a given distance along a slope. I have 
only one point on my new slope and no endpoints known. Any 

   end if

Date: 08/20/99 at 13:07:43
From: Doctor Peterson
Subject: Re: Cad Geometry

Hi, Tom.

Your specific question, stated in the simplest terms, is this: you 
have a point P = (a,b) on a line, and the slope m of the line, and you 
want to find the point Q a given distance d along the line from that 
point P. Let's draw the situation:

           |                          /
           |                         /
           |                        +Q
           |                       /
           |                     d/
           |                     /
           |                    +P (a,b)
           |                 / /
           |              /   /
           |           /     /
           |        /       /slope=m
           |     /         /
           |  /           /
           O            /
           |           /
           |          /

The most natural way to approach this is with vectors. The point Q is 
the sum of the vector OP, which you know, and the vector PQ, for which 
you know the slope and the length. Here's how we can find PQ, which 
we'll call D:

First, we can find a vector V in the direction of the slope; let's use 
V = (1, m). (This means we go over 1 and up m, producing the right 

Second, we have to find a vector in the same direction, whose length 
is 1. Since the length of our vector V is sqrt(1 + m^2), if we divide 
by this we will get a unit vector:

               1            m
     U = (-----------, -----------)
          sqrt(1+m^2)  sqrt(1+m^2)

Now we can multiply this by the distance d to get a vector with the 
right direction and length:

               d           dm
     D = (-----------, -----------)
          sqrt(1+m^2)  sqrt(1+m^2)

Now the point you are looking for is the sum of the vectors P and D:

                           d               dm
     Q = P + D = (a + -----------, b + -----------)
                      sqrt(1+m^2)      sqrt(1+m^2)

If you aren't familiar with vectors, you should be, because they make 
graphics work much easier.

Back to your original problem, you may want to note that you don't 
need any trig at all. Since CenDist = cos(Ang) and Ang is an arccos, 
you can get CenDist directly as

     CenDist = ((DistC^2)+(Rad1^2)-(Rad2^2))/(2*DistC*Rad1) * Rad1

Dist is one leg of a right triangle with hypotenuse Rad1 and other leg 
CenDist, so you can use the Pythagorean Theorem to get it. Other than 
that, the rest of your work looks fine. (Don't forget, of course, that 
there will be two points of intersection in opposite directions.)

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Linear Algebra

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