Basic Geometric Modeling Tools (Basic Computer Graphics) Part 4

Circle Formulas

Formula. Letbe noncollinear points inand The unique circle that contains the points pi has centerwhere

and radius r, where

(a) Region subtended by curve

(b) Approximation to ith interval

Figure 6.20. Circle through three points.

In the special case whereare    points    in    the    plane,    these    formulas    can be simplified. LetThen

Proof. We shall give essentially two proofs for this formula. We give a geometric argument in the general case and an algebraic one in the planar case. In either case, the solution to this problem will be easier if we first move the point p1 to the origin and solve the problem of finding the center c of the circle through the points 0, a, and b. Consider the following planes and their point-normal equations:

plane	equation
the plane X containing the points 0, a, and b (and c)
the plane X1 that is the perpendicular bisector of the segment [0,a]
the plane X2 that is the perpendicular bisector of the segment [0,b]

Basic geometric facts about circles imply that c is the intersection of these three planes X, X1, and X2. See Figure 6.20 where we have identified X with the xy-plane. Letting x be 0 in the equation for X and c in the other two equations gives us that

Applying Formula 6.5.6 for the intersection of three planes now gives us our formula (6.32). The formula for the radius r is gotten by substituting formula (6.32) into the equationand    simplifying.

Next, assume that our points pi lie in the xy-plane. The equation for the circle with center c and radius r is

Sinceequation    (6.34)    reduces    to

Substituting points a and b into equation (6.35) gives two equations in two unknowns c1 and c2:

The solution to this system of equations leads to the formula (6.33). This finished the proof of Formula 6.8.1.

The next two formulas are useful in blending computations.

Formula. Let L1 and L2 be intersecting lines in the plane defined by equations

respectively. The circles of radius r that are tangent to L1 and L2 have centers (d,e) defined by

Proof. From Figure 6.21(a) one can see that there are four solutions in general. Let L1/ and L2′ be lines that are parallel to and a distance r from lines L1 and L2, respectively. See Figure 6.21(b). There are four such pairs of lines and it is easy to see that the intersection of these lines defines the centers (d,e) of the circles we seek. Now the lines L1/ and L2′ are a distance r closer or further to the origin than the lines L1 and L2. Therefore, it is easy to see from equation (6.24) that the equations for L1/ and L2′ are

Solving equations (6.37) gives our answer.

Figure 6.21. Circles of fixed radius tangent to two lines.

Figure 6.22. Lines through a point tangent to a circle.

Formula. Let A be a point and C a circle with center B and radius r. Assume that |AB| > r. There are two lines through A that are tangent to C and they intersect C in the points D± defined by

where u and v are the orthonormal vectors

Proof. Figure 6.22(a) shows the two lines L and L’ that pass through A and are tangent to C. By switching to the coordinate system defined by the frame (A,u,v) we may assume that A = (0,0) and B = (b,0). Let D = (d1,d2). See Figure 6.22(b). The following equations are satisfied by D:

In other words,

If follows that

Since b corresponds to |AB| in the original problem, our solution translates into the stated one in world coordinates.

Formula. Consider two circles in the plane centered at points A and B with radii r1 and r2, respectively. Assume thatLetbe

the points where the four lines Li that are tangent to both of these circles intersect the circles. See Figure 6.23(a). Then

whereand    u    and    v    are the orthonormal vectors

Figure 6.23. Lines tangent to two circles.

Proof. Note that the associated pairs of linesintersect    on    the    line through the points A and B. We again switch to the coordinate system defined by the frame (A,u,v) and assume thatSee Figure 6.23(b). Let

be the point where L1 intersects the x-axis. Then

Because the triangles ADF and BEF are similar, we get that

Case 1:In    this    case    equation    (6.39)    implies    that

Case 2:In    this    case    equation    (6.38)    implies    that

In either case, since we now know F, we can now use Formula 6.8.3 to compute For example, in Case 1,

Note that to use Formula 6.8.3 we must use the frameIt is now a simple matter to rewrite this formula in the form stated earlier.

Formula. Consider two circles in the plane centered at points A and B with radii r1 and r2, respectively. Assume thatThe circles of

radius r that are tangent to these circles have center C defined by

where

Figure 6.24. Circles of fixed radius tangent to two circles.

Proof. We can see from Figure 6.24(a) that there are precisely two circles of radius r which are tangent to both circles. What we have to do is find the intersection of two circles: one has center A and radiusand the other has center B and radius

We switch to the coordinate system defined by the frame (A,u,v). Let A = (0,0) and B = (b,0). See Figure 6.24(b). We must solve the equations

We get that

This translates to the desired formula in world coordinates since b corresponds to |AB|.

This concludes our list of formulas involving circles. Other formulas can be found in [Chas78] and [BowW83].