Chapter 10Conic Sections and

Systems of Nonlinear Equations

§ 10.1

Distance and Midpoint Formulas; Circles

Blitzer, Intermediate Algebra, 5e – Slide #3 Section 10.1

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, Intermediate Algebra, 5e – Slide #4 Section 10.1

Distance Formula & Midpoint Formula

The Distance FormulaThe distance, d, between the points and in the rectangular coordinate system is

11, yx 22 , yx

.212

212 yyxxd

The Midpoint FormulaConsider a line segment whose endpoints are and

. The coordinates of the segment’s midpoint are

To find the midpoint, take the average of the two x-coordinates and the average of the two y-coordinates.

11, yx

22 , yx

.2

,2

2121

yyxx

Blitzer, Intermediate Algebra, 5e – Slide #5 Section 10.1

Distance Formula & Midpoint Formula

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

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.

(a) Letting and , we obtain 1,4, 11 yx 3,2, 22 yx

212

212 yyxxd Use the distance formula.

22 1342 Substitute the given values.

22 26 Perform subtractions within the grouping symbols.

436 Square 6 and -2.

32.640 Add and simplify.

Blitzer, Intermediate Algebra, 5e – Slide #6 Section 10.1

Distance Formula & Midpoint Formula

CONTINUECONTINUEDD 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.

2,12

4,

2

2

2

31,

2

24Midpoint

The midpoint of the line segment between the endpoints is (-1,-2).

Blitzer, Intermediate Algebra, 5e – Slide #7 Section 10.1

Equation of a Circle

Definition of a CircleA 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

.222 rkyhx

Blitzer, Intermediate Algebra, 5e – Slide #8 Section 10.1

Equation of a Circle

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Write the standard form of the equation of the circle with center (-3, 5) and radius 3.

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 equation.

222 rkyhx

Substitute -3 for h, 5 for k and 3 for r. 222 353 yx

Simplify. 953 22 yx

The standard form of the equation of the circle is .953 22 yx

Blitzer, Intermediate Algebra, 5e – Slide #9 Section 10.1

Equation of a Circle

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the center and radius of the circle whose equation is

and graph the equation.

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,

423 22 yx

.222 rkyhx 423 22 yx

222 223 yx

This is with h = -3.

2hx This is with k = 2.

2ky

This is with r = 2.

2r

Blitzer, Intermediate Algebra, 5e – Slide #10 Section 10.1

Equation of a Circle

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.

CONTINUECONTINUEDD

Blitzer, Intermediate Algebra, 5e – Slide #11 Section 10.1

Equation of a Circle

CONTINUECONTINUEDD

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5

(-3,2)

(-3,0)

(-5,2)

(-3,4)

(-1,2)

This circle is said to be “tangent” to the x-axis at (-3,0), for the circle “kisses” the x-axis there.

Blitzer, Intermediate Algebra, 5e – Slide #12 Section 10.1

Equation of a Circle

The General Form of the Equation of a Circle

The general form of the equation of a circle is

.022 FEyDxyx

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, Intermediate Algebra, 5e – Slide #13 Section 10.1

Equation of a Circle

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Write in standard form and graph:

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.

.0912422 yxyx

0912422 yxyx This is the given equation.

9124 22 yyxx Rewrite in anticipation of completing the square.

3649361244 22 yyxx Complete the square on x and on y.

Factor on the left and add on the right.

4962 22 yx

Blitzer, Intermediate Algebra, 5e – Slide #14 Section 10.1

Equation of a Circle

222 762 yx

CONTINUECONTINUEDD This last equation is in standard form. We can identify the circl

e’s center and radius by comparing this equation to the standard form of the equation of a circle, .222 rkyhx

This is with h = 2.

2hx This is with k = 6.

2ky

This is with r = 7.

2r

We use the center, (h, k) = (2, 6), and the radius, r = 7, to graph the circle. The graph is shown below.

4962 22 yx

Blitzer, Intermediate Algebra, 5e – Slide #15 Section 10.1

Equation of a Circle

CONTINUECONTINUEDD

-15

-12

-9

-6

-3

0

3

6

9

12

15

-15 -12 -9 -6 -3 0 3 6 9 12 15

(2,6)

7

7

7

7

Blitzer, Intermediate Algebra, 5e – Slide #16 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, Intermediate Algebra, 5e – Slide #17 Section 10.1

Distance Formula & Midpoint Formula

Now….Maybe you should

just take a break, go outside,

and find some circles in nature.

It’s a beautiful world…