Analytic Geometry

Section 2-6: Circles

Objective: To find equations of

circles and to find the coordinates of any points where circles and lines

meet.

Page 81

Definition of a Circle

• A circle is the set of all points in a plane

equidistant from a fixed point called the

center point.

• We can derive the equation directly

from the distance formula.

• If we place the center point on the origin

point, the equation of a circle with

center point (0, 0) and radius r is:

• x2 + y2 = r2

Circles and Points of Intersection circle: in a plane, the set of points

equidistant from a given point, called the

center.

radius: any segment whose endpoints are

the center and a point on the circle.

If the circle is centered at (0, 0), and the

radius is r, then the distance to any

point, (x, y) on the circle (using the

distance formula) is

x 0 2 y 0

2 r

x 0 2 y 0

2 r

2

x2 y

2 r

2

square both sides...

now simplify...

This is the standard form of a circle with center (0,0) and

radius r.

(x,y)

Example: Write an equation of the circle with its

center at the origin with the point (–3, 4) on the

circle.

Use the Standard Equation of the

Circle to find the radius of the

circle... x2 y

2 r

2

3 2 4

2 r

2

9 16 r2

25 r2

x2 y

2 25

(x - h)2 + (y - k)2 = r2

With center at (h, k) and radius r.

Here are some examples:

Sample Problems

1) Find the center, radius and graph the equation: (x - 2)2 + (y + 5)2 = 17

Solution: Center point : (2, -5), radius =

Find the center and radius of the circle with the

following equation: 100x2 + 100y2 – 100x + 240y – 56 = 0.

This is the given equation.

Move the loose number

over to the other side.

Group the x-stuff and y-stuff

together.

Divide off by whatever is

multiplied on the squared

terms.

Find the center and radius of the circle with the

following equation: 100x2 + 100y2 – 100x + 240y – 56 = 0.

Take the coefficient on the x-term,

multiply by one-half, square,

and add inside the x-stuff and

also to the other side. Do the

same with the y-term.

Convert the left-hand side to

squared form, and simplify the

right-hand side.

If necessary, fiddle with signs and

exponents to make your

equation match the circle

equation's format.

Find the center and radius of the circle with the

following equation: 100x2 + 100y2 – 100x + 240y – 56 = 0.

Read off the answer.

The center is at ( 1/2, – 6/5 )

and

the radius is 3/2.

2) x2 + y2 - 8x + 4y - 8 = 0 Find center, radius and graph.

Solution: We need to put the equation into the correct

form. We will do this by completing the square!!

(x2 - 8x ) + (y2 + 4y ) = 8 Complete the square!

See 1.6

(x2 - 8x + 16) + (y2 + 4y + 4) = 8 + 16 + 4

(x - 4)2 + (y + 2)2 = 28 Now it's in the correct form!!

Center point (4, -2) with radius =

4) Sketch the graph of

Solution: The above graph is part of a circle. Why?

3) Find the intersection of the line y = x - 1 and the circle x2 +

y2 = 25.

Solution: x2 + (x - 1)2 = 25

x2 + x2 - 2x + 1 = 25

2x2 - 2x - 24 = 0

x2 - x - 12 = 0 Now factor

(x - 4)(x + 3) = 0

x = 4 or x = -3

Substituting back in to find y gives the following

points: (4, 3) and (-3, -4)

Example: Find the points of intersection, if any, of the

graphs of x2 + y2 = 25 and y = 2/3 x + 2 Use substitution…

x2 + (2/3 x + 2)2 = 25 simplify ...

multiply by 9

13/9 x2 + 8/3 x + 4 = 25

13/9 x2 + 8/3 x – 21 = 0

13x2 + 24x – 189 = 0 factor 13

1

63

–3

63 –39 (13x + 63)(x – 3) = 0

13x + 63 = 0 or x – 3 = 0

x = –63 or x = 3

13 Now, find the y-coordinate for each x-coordinate...

y = 2/3(–63/13) + 2 y = 2/3 (3) + 2

y = –126/39 + 2 y = 2 + 2

y = –126/39 + 78/39 y = 4

y = –48/39

y = –16/13

The points of intersection are (–63/13, –16/13) and (3, 4)

x2 + 4/9 x2 + 8/3 x + 4 = 25

To find the Intersection of a Line and a Circle

Algebraically:

1. Solve the linear equation for y in terms of x (or x in terms of y).

2. Substitute this expression for y (or x) in the equation of the circle. Then solve the resulting quadratic equation.

3. Substitute each real x-solution from step 2 in the linear equation to get the corresponding value of y (or vice versa). Each point (x,y) is an intersection point.

4. You can check your result by substituting the coordinates of the intersection points in the two original equations.

Example: Find the equation of the circle that passes

through the points P(1, -2), Q(5, 4), and R(10, 5).

2 2

2 2

0

Using Points P 1,-2 , 5,4 ,and 10,5

1 4 2 0

25 16 5 4 0

100 25 10 5 0

18, 6, 25.

18 6 25 0

x y Dx Ey F

Q R

D E F

D E F

D E F

D E F

x y x y

Example A circle is tangent to the line 2x – y + 1 =

0 at the point (2, 5), and the center is on the line

x + y= 9. Find the equation of the circle.

8

6

4

2

-2

-4

-6

-8

-15 -10 -5 5 10 15

(2,5)

(6,3)

B

The line through (2, 5) and perpendicular to the

line 2x – y + 1 = 0 passes through the center of

the circle. The equation of this line is x + 2y =

12.

Hence, the solution of the system

X + 2y = 12

X + y = 9

Yields the coordinates of the center.

Accordingly, the center is at (6, 3). The distance

from this point to (2, 5) is √20. The equation of

the circle, therefore, is

(x – 6)2 + (y – 3)2 = 20.

Example A triangle has its sides on the lines x + 2y –

5 = 0, 2x – y – 10 = 0, and 2x + y + 2 = 0. Find the

equation of the circle inscribed in the triangle. 8

6

4

2

-2

-4

-6

-8

-15 -10 -5 5 10 15

d3

d2

d1

p

2 2 2 2

2 ' ' 2 2 ' ' 10

2 1 2 ( 1)

2 ' ' 2 2 ' ' 10

4 ' 8 0

4 ' 8

' 2

Bisector

x y x y

x y x y

x

x

x

2 2 22

2 ' ' 10 ' 2 ' 5

1 22 1

2 ' ' 10 ' 2 ' 5

' 3 ' 5 0

' 3 ' 5

Bisector

x y x y

x y x y

x y

x y

2 2

' 2

'-3 ' 5

2 -3 ' 5

-3 ' 3

' -1

Intersection point of the bisectors

is (2, -1) This is the center of the circle.

The distance from the center to the side of the circle is 5.

2 1 5.

x

x y

y

y

y

x y

Example: Find the equation of

a circle that passes through the

points (2,-1), (0,2), and (1,1).

Example: Find the equation of

the circle passing through the

points (1,-2) and (5,3) and

having its center on the line

x – y + 2 =0.

Example: A circle is tangent to

the line 2x – y + 1 = 0 at the

point (2,5), and the center of the

circle falls on the line x + y = 9.

What is the equation of the

circle?

Example: Find the equation of

the circle inscribed in the

triangle determined by the lines,

L1: 2x – 3y + 21=0,

L2: 3x – 2y – 6 = 0,

L3: 2x + 3y + 9 = 0.

Example: Derive the equation

of the circle circumscribing the

triangle determined by the lines,

x + y = 8,

2x + y = 14,

3x + y = 22.

Example: Derive the equation

of the circle which passes

through the point (-2, 1) and is

tangent to the line

3x – 2y – 6 = 0

at the point (4,3).

Homework

Assignment

Page 87- 89

Problems 1 – 49 odd