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PRACTICE PROBLEMS: Find the coordinates of the center and the radius for the
circles described by the following equations:
ANSWERS:
In certain situations you will want to consider the
following general form of a circle
as the equation of a circle in which the specific values
of the constants B, C, and D are to be determined. In this problem the unknowns
to be found are not x and y, but the values of the constants B, C, and D. The
conditions that define the circle are used to form algebraic relationships
between these constants. For example, if one of the conditions imposed on the
circle is that it pass through the point (3,4), then the general form is
written with x and y replaced by 3 and 4, respectively;
thus,
is rewritten as
Three independent constants (B, C, and D) are in the
equation of a circle; therefore, three conditions must be given to define a
circle. Each of these conditions will yield an equation with B, C, and D as the
unknowns. These three equations are then solved simultaneously to determine the
values of the constants, which satisfy all of the equations. In an analysis,
the number of independent constants in the general equation of a curve indicate
how many conditions must be set before a curve can be completely defined. Also,
the number of unknowns in an equation indicates the number of equations that
must be solved simultaneously to find the values of the unknowns. For example,
if B, C, and D are unknowns in an equation, three separate
equations involving these variables are required for a solution. A
circle may be defined by three noncollinear points; that is, by three points
not lying on a straight line. Only one circle is possible through any three
noncollinear points. To find the equation of the circle determined by three points,
substitute the x and y values of each of the given points into the general
equation to form three equations with B, C, and D as the unknowns. These
equations are then solved simultaneously to find the values of B, C and D in
the equation which satisfies the three given conditions. The
solution of simultaneous equations involving two variables is discussed in Mathematics, Volume 1. Systems involving
three variables use an extension of the same principles, but with three
equations instead of two. Step-by-step explanations of the solution are given
in the example problems. EXAMPLE: Write the equation of the circle that passes through the points (2,8),
(5,7), and (6,6). SOLUTION.- The method used in this solution corresponds to the
addition-subtraction method used for solution of equations involving two
variables. However, the method or combination of methods used depends on the
particular problem. No single method is best suited to all problems. First,
write the general form of a circle:
For
each of the given points, substitute the given values for x and y and rearrange
the terms:
To
aid in the explanation, we number the three resulting equations:
The first step is to eliminate one of the unknowns and
have two equations and two unknowns remaining. The coefficient of D is the same
in all three equations and is, therefore, the one most easily eliminated by
addition and subtraction. To eliminate D, subtract
equation (2) from equation (1):
We now have two equations, (4) and (5), in two unknowns
that can be solved simultaneously. Since the coefficient of C is the same in
both equations, it is the most easily eliminated variable. To eliminate C, subtract equation (4) from equation (5):
To find the value of C, substitute the value found for B
in equation (6) in equation (4) or (5)
Now the values of B and C can be substituted in any one of
the original equations to determine the value of D. If the values are substituted in equation (1),
The solution of the system of equations gave values for
three independent constants in the general equation
When the constant values are substituted, the equation
takes the form of
Now rearrange and complete the square in both x and y:
The equation now corresponds to a circle with its center
at (2,3) and a radius of 5. This is the circle passing through three given
points, as shown in figure 2-7, view A. The previous example problem showed one method we can use
to determine the equation of a circle when three points are given. The next
example shows another method we can use to solve the same problem. One of the
most important things to keep in mind when you study analytic geometry is that
many problems may be solved by more than one method. Each problem should be
analyzed carefully to determine what relationships exist between the given data
and the desired results of the problem. Relationships such as distance from one
point to another, distance from a point to a line, slope of a line, and the
Pythagorean theorem will be used to solve various problems.
Figure 2-7.-Circle described by three points. EXAMPLE: Find the equation of the circle that passes through
the points (2,8), (5,7), and (6,6). Use a method other than that used in the
previous example problem. SOLUTION: A different method of solving this problem
results from the reasoning in the following paragraphs: The center of the desired circle will be the intersection
of the perpendicular bisectors of the chords connecting points (2,8) with (5,7)
and (5,7) with (6,6), as shown in figure 2-7, view B. The perpendicular bisector of the line connecting two
points is the locus of all points equidistant from the two points. Using this
analysis, we can get the equations of the perpendicular bisectors of the two
lines. Equating the distance formulas that describe the distances
from the center, point (x,y), which is equidistant from the points (2,8) and
(5,7), gives
Squaring both sides gives
or
Canceling and combining terms results in
or
Follow the same procedure for the points (5,7) and (6,6):
Squaring each side gives
or
Canceling and combining terms gives a second equation in x
and y: 2x-2y= - 2 or
Solving the equations simultaneously gives the coordinates
of the intersection of the two perpendicular bisectors; this intersection is
the center of the circle.
Substitute the value x = 2 in one of the equations to find
the value of y:
Thus, the center of the circle is the point (2,3). The radius is the distance between the center (2,3) and
one of the three given points. Using point (2,8), we obtain
The equation of this circle is
as was found in the previous example. If a circle is to be defined by three points, the points
must be noncollinear. In some cases the three points are obviously noncollinear. Such is the case with the points (1, 1), ( - 2,2), and (- 1, -
1), since these points cannot be connected by a straight line. However, in many
cases you may find difficulty determining by inspection whether or not the
points are collinear; EXAMPLE: Find the equation of the circle that passes through
the points (1, 1), (2,2), and (3,3). SOLUTION: Substitute the given values of x and y in the
general form of the equation of a circle to get three equations in three
unknowns:
To eliminate D, first
subtract equation (1) from equation (2):
Then subtract equation (5) from equation (4) to eliminate
one of the unknowns:
This solution is not valid, so no circle passes through
the three given points. You should attempt to solve equations (4) and (5) by
the substitution method. When the three given points are collinear, an
inconsistent solution of some type will result. If you try to solve the problem by eliminating both B and
C at the same time (to find D), another type of inconsistent solution results.
With the given coefficients you can easily eliminate both A and B at the
same time. First, multiply equation (2) by 3 and
equation (3) by - 2 and add the resultant equations:
Then multiply equation (1) by - 2 and add
the resultant to equation (2):
This gives two values for D, which is
inconsistent since each of the constants must have a unique value consistent
with the given conditions. The three points are on the straight line y = x. PRACTICE PROBLEMS: In each of the following problems, find the equation of
the circle that passes through the three given points:
ANSWERS:
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