11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (2024)

Table of Contents
Learning Objectives Be Prepared 11.1 Be Prepared 11.2 Be Prepared 11.3 Use the Distance Formula Example 11.1 Solution Try It 11.1 Try It 11.2 Distance Formula Example 11.2 Solution Try It 11.3 Try It 11.4 Example 11.3 Solution Try It 11.5 Try It 11.6 Use the Midpoint Formula Midpoint Formula Example 11.4 Solution Try It 11.7 Try It 11.8 Write the Equation of a Circle in Standard Form Circle Standard Form of the Equation a Circle Example 11.5 Solution Try It 11.9 Try It 11.10 Example 11.6 Solution Try It 11.11 Try It 11.12 Example 11.7 Solution Try It 11.13 Try It 11.14 Graph a Circle Example 11.8 Solution Try It 11.15 Try It 11.16 Example 11.9 Solution Try It 11.17 Try It 11.18 General Form of the Equation of a Circle Example 11.10 Solution Try It 11.19 Try It 11.20 Example 11.11 Solution Try It 11.21 Try It 11.22 Media Section 11.1 Exercises Practice Makes Perfect Writing Exercises Self Check FAQs References

Learning Objectives

By the end of this section, you will be able to:

  • Use the Distance Formula
  • Use the Midpoint Formula
  • Write the equation of a circle in standard form
  • Graph a circle

Be Prepared 11.1

Before you get started, take this readiness quiz.

Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches.
If you missed this problem, review Example 2.34.

Be Prepared 11.2

Factor: x2−18x+81.x2−18x+81.
If you missed this problem, review Example 6.24.

Be Prepared 11.3

Solve by completing the square: x2−12x−12=0.x2−12x−12=0.
If you missed this problem, review Example 9.22.

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (1)

There are four conics—the circle, parabola, ellipse, and hyperbola. The next figure shows how the plane intersecting the double cone results in each curve.

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (2)

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

Use the Distance Formula

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example 11.1

Use the rectangular coordinate system to find the distance between the points (6,4)(6,4) and (2,1).(2,1).

Solution

Plot the two points. Connect the two points
with a line.
Draw a right triangle as if you were going to
find slope.		11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (3)
Find the length of each leg.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (4)
Use the Pythagorean Theorem to find d, the
distance between the two points.		a2+b2=c2a2+b2=c2
Substitute in the values.32+42=d232+42=d2
Simplify.9+16=d29+16=d2
25=d225=d2
Use the Square Root Property.d=5d=−5d=5d=−5
Since distance, d is positive, we can eliminate
d=−5.d=−5.		The distance between the points (6,4)(6,4) and
(2,1)(2,1) is 5.

Try It 11.1

Use the rectangular coordinate system to find the distance between the points (6,1)(6,1) and (2,−2).(2,−2).

Try It 11.2

Use the rectangular coordinate system to find the distance between the points (5,3)(5,3) and (−3,−3).(−3,−3).

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (5)

The method we used in the last example leads us to the formula to find the distance between the two points (x1,y1)(x1,y1) and (x2,y2).(x2,y2).

When we found the length of the horizontal leg we subtracted 6−26−2 which is x2−x1.x2−x1.

When we found the length of the vertical leg we subtracted 4−14−1 which is y2−y1.y2−y1.

If the triangle had been in a different position, we may have subtracted x1−x2x1−x2 or y1−y2.y1−y2. The expressions x2−x1x2−x1 and x1−x2x1−x2 vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say |x2−x1||x2−x1| and |y2−y1|.|y2−y1|.

In the Pythagorean Theorem, we substitute the general expressions |x2−x1||x2−x1| and |y2−y1||y2−y1| rather than the numbers.

a2+b2=c2a2+b2=c2
Substitute in the values.(|x2−x1|)2+(|y2−y1|)2=d2(|x2−x1|)2+(|y2−y1|)2=d2
Squaring the expressions makes them positive, so we eliminate the absolute value bars.(x2−x1)2+(y2−y1)2=d2(x2−x1)2+(y2−y1)2=d2
Use the Square Root Property.d=±(x2−x1)2+(y2−y1)2d=±(x2−x1)2+(y2−y1)2
Distance is positive, so eliminate the negative value.d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2

This is the Distance Formula we use to find the distance d between the two points (x1,y1)(x1,y1) and (x2,y2).(x2,y2).

Distance Formula

The distance d between the two points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) is

d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2

Example 11.2

Use the Distance Formula to find the distance between the points (−5,−3)(−5,−3) and (7,2).(7,2).

Solution

Write the Distance Formula.d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2
Label the points, (−5,−3x1,y1),(7,2x2,y2)(−5,−3x1,y1),(7,2x2,y2) and substitute.d=(7−(−5))2+(2−(−3))2d=(7−(−5))2+(2−(−3))2
Simplify.d=122+52d=122+52
d=144+25d=144+25
d=169d=169
d=13d=13

Try It 11.3

Use the Distance Formula to find the distance between the points (−4,−5)(−4,−5) and (5,7).(5,7).

Try It 11.4

Use the Distance Formula to find the distance between the points (−2,−5)(−2,−5) and (−14,−10).(−14,−10).

Example 11.3

Use the Distance Formula to find the distance between the points (10,−4)(10,−4) and (−1,5).(−1,5). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Solution

Write the Distance Formula.d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2
Label the points, (10,−4x1,y1),(−1,5x2,y2)(10,−4x1,y1),(−1,5x2,y2) and substitute.d=(−1−10)2+(5−(−4))2d=(−1−10)2+(5−(−4))2
Simplify.d=(−11)2+92d=(−11)2+92
d=121+81d=121+81
d=202d=202
Since 202 is not a perfect square, we can leave the answer in exact form or find a decimal approximation.d=202ord≈14.2d=202ord≈14.2

Try It 11.5

Use the Distance Formula to find the distance between the points (−4,−5)(−4,−5) and (3,4).(3,4). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try It 11.6

Use the Distance Formula to find the distance between the points (−2,−5)(−2,−5) and (−3,−4).(−3,−4). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Use the Midpoint Formula

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) is

(x1+x22,y1+y22)(x1+x22,y1+y22)

To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.

Example 11.4

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are (−5,−4)(−5,−4) and (7,2).(7,2). Plot the endpoints and the midpoint on a rectangular coordinate system.

Solution

Write the Midpoint Formula.(x1+x22,y1+y22)(x1+x22,y1+y22)
Label the points, (−5,−4x1,y1),(7,2x2,y2)(−5,−4x1,y1),(7,2x2,y2)
and substitute.		(−5+72,−4+22)(−5+72,−4+22)
Simplify.(22,−22)(22,−22)
(1,−1)(1,−1)
The midpoint of the segment is the point
(1,−1).(1,−1).
Plot the endpoints and midpoint.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (6)

Try It 11.7

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are (−3,−5)(−3,−5) and (5,7).(5,7). Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.8

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are (−2,−5)(−2,−5) and (6,−1).(6,−1). Plot the endpoints and the midpoint on a rectangular coordinate system.

Both the Distance Formula and the Midpoint Formula depend on two points, (x1,y1)(x1,y1) and (x2,y2).(x2,y2). It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, it may be easier to remember the formulas.

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (7)

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (8)

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, (h,k),(h,k), and the fixed distance is called the radius, r, of the circle.

Circle

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, (h,k),(h,k), and the fixed distance is called the radius, r, of the circle.

We look at a circle in the rectangular coordinate system.
The radius is the distance from the center, (h,k),(h,k), to a
point on the circle, (x,y).(x,y).		11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (9)
To derive the equation of a circle, we can use the
distance formula with the points (h,k),(h,k), (x,y)(x,y) and the
distance, r.		d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2
Substitute the values.r=(x−h)2+(y−k)2r=(x−h)2+(y−k)2
Square both sides.r2=(x−h)2+(y−k)2r2=(x−h)2+(y−k)2

This is the standard form of the equation of a circle with center, (h,k),(h,k), and radius, r.

Standard Form of the Equation a Circle

The standard form of the equation of a circle with center, (h,k),(h,k), and radius, r, is

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (10)

Example 11.5

Write the standard form of the equation of the circle with radius 3 and center (0,0).(0,0).

Solution

Use the standard form of the equation of a circle(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
Substitute in the values r=3,h=0,r=3,h=0, and k=0.k=0.(x−0)2+(y−0)2=32(x−0)2+(y−0)2=32
11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (11)
Simplify.x2+y2=9x2+y2=9

Try It 11.9

Write the standard form of the equation of the circle with a radius of 6 and center (0,0).(0,0).

Try It 11.10

Write the standard form of the equation of the circle with a radius of 8 and center (0,0).(0,0).

In the last example, the center was (0,0).(0,0). Notice what happened to the equation. Whenever the center is (0,0),(0,0), the standard form becomes x2+y2=r2.x2+y2=r2.

Example 11.6

Write the standard form of the equation of the circle with radius 2 and center (−1,3).(−1,3).

Solution

Use the standard form of the equation of a
circle.		(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
Substitute in the values.(x−(−1))2+(y−3)2=22(x−(−1))2+(y−3)2=22
11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (12)
Simplify.(x+1)2+(y−3)2=4(x+1)2+(y−3)2=4

Try It 11.11

Write the standard form of the equation of the circle with a radius of 7 and center (2,−4).(2,−4).

Try It 11.12

Write the standard form of the equation of the circle with a radius of 9 and center (−3,−5).(−3,−5).

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Example 11.7

Write the standard form of the equation of the circle with center (2,4)(2,4) that also contains the point (−2,1).(−2,1).

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (13)

Solution

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center (2,4)(2,4) and point (−2,1)(−2,1)

Use the Distance Formula to find the radius.r=(x2−x1)2+(y2−y1)2r=(x2−x1)2+(y2−y1)2
Substitute the values. (2,4x1,y1),(−2,1x2,y2)(2,4x1,y1),(−2,1x2,y2)r=(−2−2)2+(1−4)2r=(−2−2)2+(1−4)2
Simplify.r=(−4)2+(−3)2r=(−4)2+(−3)2
r=16+9r=16+9
r=25r=25
r=5r=5

Now that we know the radius, r=5,r=5, and the center, (2,4),(2,4), we can use the standard form of the equation of a circle to find the equation.

Use the standard form of the equation of a circle.(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
Substitute in the values.(x−2)2+(y−4)2=52(x−2)2+(y−4)2=52
Simplify.(x−2)2+(y−4)2=25(x−2)2+(y−4)2=25

Try It 11.13

Write the standard form of the equation of the circle with center (2,1)(2,1) that also contains the point (−2,−2).(−2,−2).

Try It 11.14

Write the standard form of the equation of the circle with center (7,1)(7,1) that also contains the point (−1,−5).(−1,−5).

Graph a Circle

Any equation of the form (x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2 is the standard form of the equation of a circle with center, (h,k),(h,k), and radius, r. We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from x and y. In the next example, the equation has x+2,x+2, so we need to rewrite the addition as subtraction of a negative.

Example 11.8

Find the center and radius, then graph the circle: (x+2)2+(y−1)2=9.(x+2)2+(y−1)2=9.

Solution

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (14)
Use the standard form of the equation of a circle.
Identify the center, (h,k)(h,k) and radius, r.		11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (15)
Center: (−2,1)(−2,1) radius: 3
Graph the circle.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (16)

Try It 11.15

ⓐ Find the center and radius, then ⓑ graph the circle: (x−3)2+(y+4)2=4.(x−3)2+(y+4)2=4.

Try It 11.16

ⓐ Find the center and radius, then ⓑ graph the circle: (x−3)2+(y−1)2=16.(x−3)2+(y−1)2=16.

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of x2,y2x2,y2 to be one.

Example 11.9

Find the center and radius and then graph the circle, 4x2+4y2=64.4x2+4y2=64.

Solution

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (17)
Divide each side by 4.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (18)
Use the standard form of the equation of a circle.
Identify the center, (h,k)(h,k) and radius, r.		11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (19)
Center: (0,0)(0,0) radius: 4
Graph the circle.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (20)

Try It 11.17

ⓐ Find the center and radius, then ⓑ graph the circle: 3x2+3y2=273x2+3y2=27

Try It 11.18

ⓐ Find the center and radius, then ⓑ graph the circle: 5x2+5y2=1255x2+5y2=125

If we expand the equation from Example 11.8, (x+2)2+(y−1)2=9,(x+2)2+(y−1)2=9, the equation of the circle looks very different.

(x+2)2+(y−1)2=9(x+2)2+(y−1)2=9
Square the binomials.x2+4x+4+y2−2y+1=9x2+4x+4+y2−2y+1=9
Arrange the terms in descending degree order, and get zero on the rightx2+y2+4x−2y−4=0x2+y2+4x−2y−4=0

This form of the equation is called the general form of the equation of the circle.

General Form of the Equation of a Circle

The general form of the equation of a circle is

x2+y2+ax+by+c=0x2+y2+ax+by+c=0

If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y. Then we can graph the circle using its center and radius.

Example 11.10

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2−4x−6y+4=0.x2+y2−4x−6y+4=0.

Solution

We need to rewrite this general form into standard form in order to find the center and radius.

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (21)
Group the x-terms and y-terms.
Collect the constants on the right side.		11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (22)
Complete the squares.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (23)
Rewrite as binomial squares.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (24)
Identify the center and radius.Center: (2,3)(2,3) radius: 3
Graph the circle.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (25)

Try It 11.19

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2−6x−8y+9=0.x2+y2−6x−8y+9=0.

Try It 11.20

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2+6x−2y+1=0.x2+y2+6x−2y+1=0.

In the next example, there is a y-term and a y2y2-term. But notice that there is no x-term, only an x2x2-term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.

Example 11.11

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2+8y=0.x2+y2+8y=0.

Solution

We need to rewrite this general form into standard form in order to find the center and radius.

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (26)
Group the x-terms and y-terms.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (27)
There are no constants to collect on the
right side.
Complete the square for y2+8y.y2+8y.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (28)
Rewrite as binomial squares.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (29)
Identify the center and radius.Center: (0,−4)(0,−4) radius: 4
Graph the circle.11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (30)

Try It 11.21

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2−2x−3=0.x2+y2−2x−3=0.

Try It 11.22

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2−12y+11=0.x2+y2−12y+11=0.

Media

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

  • Distance-Midpoint Formulas and Circles
  • Finding the Distance and Midpoint Between Two Points
  • Completing the Square to Write Equation in Standard Form of a Circle

Section 11.1 Exercises

Practice Makes Perfect

Use the Distance Formula

In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

1.

(2,0)(2,0) and (5,4)(5,4)

2.

(−4,−3)(−4,−3) and (2,5)(2,5)

3.

(−4,−3)(−4,−3) and (8,2)(8,2)

4.

(−7,−3)(−7,−3) and (8,5)(8,5)

5.

(−1,4)(−1,4) and (2,0)(2,0)

6.

(−1,3)(−1,3) and (5,−5)(5,−5)

7.

(1,−4)(1,−4) and (6,8)(6,8)

8.

(−8,−2)(−8,−2) and (7,6)(7,6)

9.

(−3,−5)(−3,−5) and (0,1)(0,1)

10.

(−1,−2)(−1,−2) and (−3,4)(−3,4)

11.

(3,−1)(3,−1) and (1,7)(1,7)

12.

(−4,−5)(−4,−5) and (7,4)(7,4)

Use the Midpoint Formula

In the following exercises, ⓐ find the midpoint of the line segments whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.

13.

(0,−5)(0,−5) and (4,−3)(4,−3)

14.

(−2,−6)(−2,−6) and (6,−2)(6,−2)

15.

(3,−1)(3,−1) and (4,−2)(4,−2)

16.

(−3,−3)(−3,−3) and (6,−1)(6,−1)

Write the Equation of a Circle in Standard Form

In the following exercises, write the standard form of the equation of the circle with the given radius and center (0,0).(0,0).

17.

Radius: 7

18.

Radius: 9

19.

Radius: 22

20.

Radius: 55

In the following exercises, write the standard form of the equation of the circle with the given radius and center

21.

Radius: 1, center: (3,5)(3,5)

22.

Radius: 10, center: (−2,6)(−2,6)

23.

Radius: 2.5,2.5, center: (1.5,−3.5)(1.5,−3.5)

24.

Radius: 1.5,1.5, center: (−5.5,−6.5)(−5.5,−6.5)

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

25.

Center (3,−2)(3,−2) with point (3,6)(3,6)

26.

Center (6,−6)(6,−6) with point (2,−3)(2,−3)

27.

Center (4,4)(4,4) with point (2,2)(2,2)

28.

Center (−5,6)(−5,6) with point (−2,3)(−2,3)

Graph a Circle

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

29.

( x + 5 ) 2 + ( y + 3 ) 2 = 1 ( x + 5 ) 2 + ( y + 3 ) 2 = 1

30.

( x − 2 ) 2 + ( y − 3 ) 2 = 9 ( x − 2 ) 2 + ( y − 3 ) 2 = 9

31.

( x − 4 ) 2 + ( y + 2 ) 2 = 16 ( x − 4 ) 2 + ( y + 2 ) 2 = 16

32.

( x + 2 ) 2 + ( y − 5 ) 2 = 4 ( x + 2 ) 2 + ( y − 5 ) 2 = 4

33.

x 2 + ( y + 2 ) 2 = 25 x 2 + ( y + 2 ) 2 = 25

34.

( x − 1 ) 2 + y 2 = 36 ( x − 1 ) 2 + y 2 = 36

35.

( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25 ( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25

36.

( x − 1 ) 2 + ( y − 3 ) 2 = 9 4 ( x − 1 ) 2 + ( y − 3 ) 2 = 9 4

37.

x 2 + y 2 = 64 x 2 + y 2 = 64

38.

x 2 + y 2 = 49 x 2 + y 2 = 49

39.

2 x 2 + 2 y 2 = 8 2 x 2 + 2 y 2 = 8

40.

6 x 2 + 6 y 2 = 216 6 x 2 + 6 y 2 = 216

In the following exercises, ⓐ identify the center and radius and ⓑ graph.

41.

x 2 + y 2 + 2 x + 6 y + 9 = 0 x 2 + y 2 + 2 x + 6 y + 9 = 0

42.

x 2 + y 2 − 6 x − 8 y = 0 x 2 + y 2 − 6 x − 8 y = 0

43.

x 2 + y 2 − 4 x + 10 y − 7 = 0 x 2 + y 2 − 4 x + 10 y − 7 = 0

44.

x 2 + y 2 + 12 x − 14 y + 21 = 0 x 2 + y 2 + 12 x − 14 y + 21 = 0

45.

x 2 + y 2 + 6 y + 5 = 0 x 2 + y 2 + 6 y + 5 = 0

46.

x 2 + y 2 − 10 y = 0 x 2 + y 2 − 10 y = 0

47.

x 2 + y 2 + 4 x = 0 x 2 + y 2 + 4 x = 0

48.

x 2 + y 2 − 14 x + 13 = 0 x 2 + y 2 − 14 x + 13 = 0

Writing Exercises

49.

Explain the relationship between the distance formula and the equation of a circle.

50.

Is a circle a function? Explain why or why not.

51.

In your own words, state the definition of a circle.

52.

In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (31)

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

11.1 Distance and Midpoint Formulas; Circles - Intermediate Algebra 2e | OpenStax (2024)

FAQs

What is the formula for finding the midpoint of a circle? ›

Assuming you have either endpoint of the diameter of a circle, you can use the midpoint formula to find the point midway between the two points. According to the definition of a diameter, this will be the circle's center point. If you have the points (x1,y1) and (x2,y2) , the midpoint formula is (x1+x22,y1+y22) .

Read On ›
How to find the center of a circle using the midpoint formula site 1? ›

The center of a circle is the midpoint of the diameter. So, by using the midpoint formula, if the endpoints of the diameter are (a, b) and (c, d), then the coordinates of the center of circle are [(a + c)/2, (b + d)/2].

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What is the distance formula of a circle? ›

Definition 11.2. 3
We look at a circle in the rectangular coordinate system. The radius is the distance from the center, (h,k),to a point on the circle, (x,y).Figure 11.1.11
To derive the equation of a circle, we can use the distance formula with the points (h,k), (x,y) and the distance, r.d=√(x2−x1)2+(y2−y1)2
2 more rows
Feb 13, 2022

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How to solve midpoint formula? ›

The midpoint formula is just an average. Add the 2 X-values, then divide by 2. Add the 2 Y-values, then divide by 2. You have then found the average for the X and Y values which gives you the point half way between the original 2 points.

Learn More ›
How to find the center of a circle in Algebra 2? ›

In order to find the center and radius, we need to change the equation of the circle into standard form, ( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2+(y-k)^2=r^2 (x−h)2​+(y−k)2​=r2​, where h and k are the coordinates of the center and r is the radius.

Read The Full Story ›
What is the formula for midpoint and distance? ›

The midpoint formula is M = ( x 1 + x 2 2 , y 1 + y 2 2 ) ), where M is the midpoint and the other terms have the same definitions as they do in the distance formula. The slope equation is m = y 2 − y 1 x 2 − x 1 , where m is the slope and the other terms have the same definitions as they do in the distance formula.

View More ›
What is the formula for finding the midpoint answer? ›

How do you Calculate Midpoint? The midpoint can be found with the formula ((x1 + x2)/2, (y1 + y2)/2). Here (x1, y1), and (x2, y2) are the coordinates of two points, and the midpoint is a point lying equidistant and between these two points.

See Details ›
What is the equation of a circle? ›

Explanation: The formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.

Get More Info ›
What is half of a circle? ›

A semicircle is simply half of a circle. That means if you take a circle and slice it down its diameter, or the line that runs through the circle's interior and includes its midpoint, you'll end up with two semicircles.

Explore More ›
What are all the formulas for a circle? ›

Circle Formulas
ParametersCircle Formulas
Diameter of a circle formulaD = 2 × r
Circumference of a circle formulaC = 2 × π ×
Area of a circle formulaA = π × r2
Sep 11, 2023

Read On ›

What is the general formula of a circle? ›

The general form of the equation of circle is: x2 + y2 + 2gx + 2fy + c = 0. This general form is used to find the coordinates of the center of the circle and the radius of the circle. Here, c is a constant term, and the equation having c value represents a circle that is not passing through the origin.

View Details ›
What is the rule of midpoint formula? ›

The midpoint can be found with the formula ((x1 + x2)/2, (y1 + y2)/2). Here (x1, y1), and (x2, y2) are the coordinates of two points, and the midpoint is a point lying equidistant and between these two points.

Show Me More ›
What is the formula for the midpoint section? ›

Special Case: Midpoint Formula

When the point P(x, y) divides the line segment into two halves, we may say that P(x, y) is the midpoint of the line segment. By the use of section and midpoint formula: m : n = 1 : 1, since the line is divided into equal parts, we have.

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How do you find the midpoint of a line in algebra? ›

This looks complicated when written algebraically, but we're basically calculating the (mean) average of both the x values and the y values. We sum the two x coordinates and divide by 2 to find the x coordinate of the midpoint, and we sum the two y coordinates and divide by 2 to find the y coordinate of the midpoint.

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Which formula is used to find the midpoint of two numbers? ›

We can find the midpoint of the two given numbers by adding both the numbers and then dividing it by the number two. It is the average of the two numbers. Add both the given numbers and then divide it by the number two.

Show Me More ›

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