Circles, Spheres, and Hyper-Spheres

This page presents the problem of finding the equation of a

circle in two space, given three non-colinear points;
sphere in three space, given four non-coplanar points;
hyper-sphere in n-space given (n+1) non-cohyperplanar points.

Circle

The general equation of a circle in two-space is

where the center of the circle is at the point (a,b) and the radius of the circle is r. Our task is to find the values of a, b, and r if we are given three non-colinear points on the circle. [Note that if they are on the circle then they cannot be colinear.]

As an explicit example, consider the points F=(3,5), G=(9,1), and H=(14,16).

We would graph the points as in Figure 1. Figure 1

Then we would find the midpoint of the segment FG, this turns out to be the point (6,3) and we would construct the perpendicular bisector of the segment FG. We woud do the same for the segment FH, finding the midpoint at (8.5,10.5), and construct the perpendicular bisector to FH. These are all shown in Figure 2. Figure 2

The point where the two perpendicular bisectors of chords of the circle intersect is the center of the circle, clearly at (10,9) in Figure 2. And a radius of the circle would be the line segment from the center to the point F, also shown in Figure 2. Using the Pythagorean Theorem, we would find that the length of the radius is 8. From this we find that the equation of the circle is

. If we construct such a circle we get the image shown in Figure 3. Figure 3

Let us return to the original problem statement and solve this via algebra. First, we know that the points F, G, and H are all points on the circle. Thus, we can substitute the coordinates of any of those points into the equation for the circle. Therefore, we have the following three equations, all of which are true:

Notice that these are three equations in 3 unknowns, a, b, and r. Expanding the three equations gives us

Again, three equations in three unknowns. Unfortunately, these are quadratic equations. We would really prefer to be dealing with linear equations. We form new equations by finding the difference between the second and first and then the difference between the third and the second equations. This gives us

These simplify to

These are two linear equations in two unknowns. We can solve this on the calculator using a matrix and the rref function as shown in Figure 4. Figure 4

This means that the value of a is 10 and the value of b is 9, meaning that the center of the circle is at (10,9). All that remains is to find the value of r. We can do that from any of the three original equations now that we know the value of a and b. Using the first equation we now have

The process that we just used,

state the general equation of a circle with center at (a,b) and radius r;
form the three quadratic equations, having unkowns a, b, and r, by using the general equation for the circle but substituting the coordinates of the three given points for the values of x and y;
expand those three equations,
form two linear equations, with unknowns a and b, by taking the difference between the second and first quadratic and then the difference between the third and the second quadratic;
put those two linear equations into standard form;
represent those two linear equations via a 2x3 matrix;
solve that matrix using rref() to obtain the values of a and b;
use any one of the original quadratic equations to solve for r.

is a general solution. That is, there was nothing special about the values of the three original points that allowed us to change the three quadratic equations into two linear equations in two unknowns.

We can restate the solution for finding the equation of a circle given three points on the circle as (x₁,y₁, (x₂,y₂, and (x₃,y₃ by stating the general form of the circle

forming the three quadratic equations

expanding the quadratic equations

forming and simplifying the two linear equations

and forming the reprentaive matrix

and then solving this via the rref() function.

Sphere

The general equation of a sphere in 3-space with center at (a,b,c), is

If we have 4 non-coplaner points in 3-space then there is a unique sphere that has those points on its surface. Or, if we know the coordinates of four points on a sphere then we should be able to find the equation of the sphere.

Although there is a geometric solution to this problem, one that follows the pattern of the geometric solution to finding the equation of a circle from the three points on the circle, it is difficult to draw such a solution. Rather, we will just expand on the algebraic solution pattern that we just saw. We can start with a specific problem. Consider the four points (3,9,2), (5,11,4), (6,1,3) and (1,1,2). To find the equation of the sphere with those points on its surface we form the four quadratic equations

We expand those equations to get

From there we form three linear equations by finding the difference of the second and first quadratic equation, the difference of the third and second quadratic equation, and the difference of the fourth and the third quadratic equation. Once simplified this becomes

We can rewrite this as a 3x4 matrix

which we can put into the calculator as

and then we can use rref() to solve that system, getting

Thus the center of the circle is at the point (2,5,10). We can get the radius by rewriting one of the original quadratic equations using the actual center values. Here we will use the second equation which becomes

Thus the radius of the circle is 9 and the final equation of the circle is

We can actually do this same problem, at least to the point of setting up the matrix of values, for the general case of having four points (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃), and (x₄,y₄,z₄). Using those points we get four quadratic equations

and these expand to

From those equations, by finding the difference between the second and the first, the third and the second, and the fourth and third, we get three linear equations

We can rewrite them as the matrix

For any four points with numeric cootdinates this resolves to a 3 by 4 numeric matrix that we can enter into the calculator and then solve via rref().

Hypersphere and the program SPHERE

Notice the pattern in both the development of our generalized matrix solution and of the variables shown in the resulting matrix. Although we may not be able to visualize moving into 4 dimensions, the pattern of our discussions, first for the circle and then for the sphere, will certainly repeat, but it will be expanded to have 5 points in 4-space producing 5 quadratic equations. Then by finding the differences between successive pairs of those 5 quadratic equations we will get 4 linear equations in 4 unknowns, which we can then represent in a 4x5 matrix that we can again solve via rref() to find the coordinates of the center of a hypersphere, the set of all points at some given distance from that center. This is the sort of pattern that we can capture in a program. The SPHERE program for our TI-83/84 calculators does this.

The program opens by displaying a title and asking for the number of dimensions in our problem:

The program keeps that value in the variable N. But if we have N then we need N+1 points, each of which has N dimensions. The program goes on to store the value N+1 in the variable P. We can use N and P to create two matrices, one to hold the coordinates of the points and one to hold the values representing the N linear equations that we would get via our process:

The matrix [A] is now a PxN matrix. As such it can hold N coordinates for each of the P points. The net part of the program cycles through each of those P points asking the user for each of the N coordinates.

At the end of that process [A] holds all of the cooordinates for all of the points.

The next task is to compute the coefficients and the constant value for each of the N linear equations. We use the pattern that we have seen in the case of the circle and the sphere to compute these. To find the coefficinet we remember that when we square the terms in the quadratic equation we get a middle term with the coefficient that looks like ^-2*w_i where w_i is the i^th coordinate of the known point. Therefore, we code subtracting the (i+1)^st equation from the i^th by using line 19 of the program to pick out the appropriate values from [A} and assign ^-2 times that difference to the appropriate location in [B]. (A small note: in the exmples from the start of the page we have subtracted the i^th> equation from the (i+1)^st, but the program does it reversed. As long as we are consistent, the change is immaterial to the final result.)

Finding the constant values for each of the linear equations, the P value in each row of [B], is just finding the sum of the differences of each of the squares of the known point coordiantes. The program does this by initializing X to 0 at line 17 and then accumulating the sum in line 20. Finally, the sum is assigned to the correct location in [B] at line 22.

Line 20 of the program performs the rref() function on the matrix [B] and stores that in matrix [C]. The final column of [C] now holds the coordinates of the center of the hypersphere. All that remains is to display the coordinates of the center and to find the radius and to do that we compute the distance from the center to one of the points. The For End construction from line 27 through line 30 not only displays the coordinates of the center, it also builds up the squares of the differences between coordinates of the first point and coordinates of the center. That accumulated sum is the square of the radius and is displayed via line 31.

Thus the program happily wrks for 2, 3, or more dimension problems.