Chapter 10
Conic Sections and Systems of
Nonlinear Equations
§ 10.1
Distance and Midpoint Formulas; Circles
Distance Formula & Midpoint Formula
We can use the Pythagorean theorem to develop a formula for finding the
distance between two points in the rectangular coordinate system. After finding a
formula for distance between two points, we can use the formula to derive a
formula for the midpoint of a line segment between two points. These two
formulas are in turn used to derive the formula for a circle.
The rectangular coordinate system gives us a unique way of knowing a circle. It
enables us to translate a circle’s geometric definition into an algebraic equation.
Circles occur everywhere in nature – in ripples on water, patterns on a butterfly’s
wings, on cross sections of trees. Some people consider the circle to be the most
pleasing of all shapes. Let’s begin by looking at the formula for the distance
between points. Look at page 732 in your text to see how the formula is derived.
Blitzer, Algebra for College Students, 6e – Slide #3 Section 10.1
Distance Formula & Midpoint Formula
The Distance Formula
The distance, d, between the points x1 , y1  and x2 , y2  in
the rectangular coordinate system is
d
x2  x1 2   y2  y1 2 .
The Midpoint Formula
Consider a line segment whose endpoints are x1 , y1  and
x2 , y2 . The coordinates of the segment’s midpoint are
 x1  x2 y1  y2 
,

.
2
2


To find the midpoint, take the average of the two xcoordinates and the average of the two y-coordinates.
Blitzer, Algebra for College Students, 6e – Slide #4 Section 10.1
Distance Formula & Midpoint Formula
EXAMPLE
Given the points (-4,-1) and (2,-3) find (a) the distance between
the points and (b) the midpoint of the line segment with the
given endpoints.
SOLUTION
(a) Letting x1, y1    4,1 and x2 , y2   2,3 , we obtain
x2  x1 2   y2  y1 2
Use the distance formula.

2   42   3  12

62   22
Substitute the given values.
Perform subtractions within the
grouping symbols.
Square 6 and -2.
Add and simplify.
d
 36  4
 40  6.32
Blitzer, Algebra for College Students, 6e – Slide #5 Section 10.1
Distance Formula & Midpoint Formula
CONTINUED
The distance between the given points is approximately 6.32
units.
(b) To find the coordinates of the midpoint, we average the
coordinates of the endpoints.
  4  2  1  3    2  4 
Midpoint  
,
,

   1,2 
2   2 2 
 2
The midpoint of the line segment between the endpoints is
(-1,-2).
Blitzer, Algebra for College Students, 6e – Slide #6 Section 10.1
Equation of a Circle
Definition of a Circle
A circle is the set of all points in a plane that are
equidistant from a fixed point, called the center. The
fixed distance from the circle’s center to any point on the
circle is called the radius.
The Standard Form of the
Equation of a Circle
The standard form of the equation of a circle with center
(h, k) and radius r is
x  h 2   y  k 2  r 2 .
Blitzer, Algebra for College Students, 6e – Slide #7 Section 10.1
Equation of a Circle
EXAMPLE
Write the standard form of the equation of the circle with center
(-3, 5) and radius 3.
SOLUTION
The center is (-3, 5). Because the center is represented as (h,k) in
the standard form of the equation, h = -3 and k = 5. The radius is
3, so we will let r = 3.
This is the standard form of a circle’s
x  h 2   y  k 2  r 2
equation.
x   32   y  52  32
Substitute -3 for h, 5 for k and 3 for r.
x  32   y  52  9
Simplify.
The standard form of the equation of the circle is x  32   y  52  9.
Blitzer, Algebra for College Students, 6e – Slide #8 Section 10.1
Equation of a Circle
EXAMPLE
Find the center and radius of the circle whose equation is
x  32   y  22  4
and graph the equation.
SOLUTION
To graph the circle, we have to know its center, (h, k), and its
radius, r. We can find the values for h, k, and r by comparing the
given equation to the standard form of the equation of a circle,
x  h 2   y  k 2  r 2 .
x  32   y  22  4
x   32   y  22  22
This is  x  h 2
with h = -3.
This is  y  k 2
with k = 2.
Blitzer, Algebra for College Students, 6e – Slide #9 Section 10.1
This is r 2
with r = 2.
Equation of a Circle
CONTINUED
We see that h = -3, k = 2, and r = 2. Thus, the circle has center
(h, k) = (-3, 2) and a radius of 2 units. To graph this circle, first
plot the center (-3, 2). Because the radius is 2, you can locate at
least four points on the circle by going out two units to the right,
to the left, up, and down from the center.
The points two units to the right and to the left of (-3, 2) are
(-1, 2) and (-5, 2), respectively. The points two units up and
down from (-3, 2) are (-3, 4) and (-3, 0), respectively.
Using these points, we obtain the graph that follows.
Blitzer, Algebra for College Students, 6e – Slide #10 Section 10.1
Equation of a Circle
CONTINUED
5
This circle is said to be “tangent”
to the x-axis at (-3,0), for the
circle “kisses” the x-axis there.
(-3,4)
4
3
(-1,2)
(-3,2)
(-5,2)
2
1
(-3,0)
0
-5
-4
-3
-2
-1
0
1
2
3
4
-1
-2
-3
-4
-5
Blitzer, Algebra for College Students, 6e – Slide #11 Section 10.1
5
Equation of a Circle
The General Form of the
Equation of a Circle
The general form of the equation of a circle is
x 2  y 2  Dx  Ey  F  0.
For the equation of a circle, there is a general form and a standard form.
In the standard form, it’s easy to see the center of the circle and it’s radius. In
this general form, we can only see that we have either a circle or some
Degenerate case of the circle. We would need to change the form to see more.
In the next example, we will do just that.
Blitzer, Algebra for College Students, 6e – Slide #12 Section 10.1
Equation of a Circle
EXAMPLE
Write in standard form and graph: x 2  y 2  4 x  12 y  9  0.
SOLUTION
Because we plan to complete the square on both x and y, let’s
rearrange the terms so that x-terms are arranged in descending
order, y-terms are arranged in descending order, and the constant
term appears on the right.
This is the given equation.
x 2  y 2  4 x  12 y  9  0
x 2  4x  y 2 12 y   9 Rewrite in anticipation of completing
the square.
x 2  4x  4 y 2 12 y  36  9  4  36 Complete the square on x
and on y.
x  22   y  62  49 Factor on the left and add on the right.
Blitzer, Algebra for College Students, 6e – Slide #13 Section 10.1
Equation of a Circle
CONTINUED
This last equation is in standard form. We can identify the
circle’s center and radius by comparing this equation to the
standard form of the equation of a circle, x  h 2   y  k 2  r 2 .
x  22   y  62  49
x  22   y  62  7 2
This is  x  h 2
with h = 2.
This is r 2
with r = 7.
This is  y  k 2
with k = 6.
We use the center, (h, k) = (2, 6), and the radius, r = 7, to graph
the circle. The graph is shown below.
Blitzer, Algebra for College Students, 6e – Slide #14 Section 10.1
Equation of a Circle
CONTINUED
15
12
7
9
7
7 6
3
(2,6)
7
0
-15 -12 -9
-6
-3 -3 0
3
6
9
12 15
-6
-9
-12
-15
Blitzer, Algebra for College Students, 6e – Slide #15 Section 10.1
In conclusion…
Some questions about circles…
A circle is the set of all points that are equidistant from a fixed point called the
center of the circle.
Question… Is the center a part of the circle? Answer – no. The circle itself is just
the set of points that are equidistant from the center.
The distance from the center to any point on the circle is the radius of the circle.
Question… Could the radius be 0? Answer – yes. In that case, what points would
make up the circle? In the case that the radius is 0, and you moved out zero from
the center point to the circle, the circle would be just the center point. This is the
only case when the center would be a part of the circle.
Question… Could the radius be a negative number? Answer – no.
Question… How do you convert from the general form of the equation of a circle to
the standard form? Answer… by completing the square.
Blitzer, Algebra for College Students, 6e – Slide #16 Section 10.1
Distance Formula & Midpoint Formula
Now….
Maybe you should
just
take a break,
go outside,
and
find some circles in nature.
Circles are everywhere… in ripples on water, patterns
on a butterfly’s wings, cross sections on trees. Some
consider the circle to be the most pleasing of all
shapes.
Blitzer, Algebra for College Students, 6e – Slide #17 Section 10.1