M11 Symmetry Parabola and Circle.pdf

Symmetry of Graphs of

Equations, Parabolas,

and Circles

At the end of this lecture, a student must be able to:

Recognize symmetry of graphs with respect to lines and points

Recognize, algebraically, symmetry of graphs with respect to the x- and y-axis and the origin and the line y=x

Symmetry of Graphs

−5−4−3−2−1 1 2 3 4 5

−5 −4 −3 −2 −1 1 2 3 4 5

0

Intuitively, a graph is

symmetric with respect to a line `

if the graph is unchanged after reflecting about the line `.

Example:

The graph ofy= 1

x is symmetric

Symmetry of Graphs

−5 −3 −1 1 3 5

−8 −6 −4 −2 2 4

0

Intuitively, a graph is

symmetric with respect to a point P if the graph is unchanged after rotating by 180◦ about the point P.

Example:

Thegraph ofy= (x+ 1)3_−x−3

Symmetry of Graphs

A graph is

symmetric with respect to the

x

-axis

if for every point (a, b) on the graph, the

point (a,

−

b) is also on the graph.

−2 −1 1 2

Symmetry of Graphs

A graph is

symmetric with respect to the

x

-axis

if for every point (a, b) on the graph, the

point

(

a,

−

b

)

is also on the graph.

−2 −1 1 2

Symmetry of Graphs

A graph is

symmetric with respect to the

y

-axis

if for every point (a, b) on the graph, the

point (

−

a, b) is also on the graph.

−2 −1 1 2

Symmetry of Graphs

A graph is

symmetric with respect to the

y

-axis

if for every point (a, b) on the graph, the

point

(−

a, b

)

is also on the graph.

−2 −1 1 2

Symmetry of Graphs

A graph is

symmetric with respect to the

origin

if for every point (a, b) on the graph, the

point (

−

a,

−

b) is also on the graph.

−2 −1 1 2

Symmetry of Graphs

A graph is

symmetric with respect to the

origin

if for every point (a, b) on the graph, the

point

(−

a,

−

b

)

is also on the graph.

−2 −1 1 2

Symmetry of Graphs

A graph is

symmetric with respect to the

line with equation y=x

if for every point (a, b)

on the graph, the point (b, a) is also on the graph.

−2 −1 1 2

Symmetry of Graphs

A graph is

symmetric with respect to the

line with equation y=x

if for every point (a, b)

on the graph, the point

(

b, a

)

is also on the graph.

−2 −1 1 2

GOAL:

Given an equation, determine whether its graph

is symmetric with respect to the coordinate axes

and the origin.

Definition

Equivalent equations

are equations that have

the same solution set.

GOAL:

Given an equation, determine whether its graph

is symmetric with respect to the coordinate axes

and the origin.

Definition

Equivalent equations

are equations that have

the same solution set.

Tests for Symmetry

1. Given an equation,if an equivalent equation is obtained when y is replaced with −y, then the graph of the equation is symmetric with respect to the

x−axis.

Ex. x=y2

x = (−y)2

x = y2

∴symmetric wrt the x−axis

−1 1 2 3 4

Tests for Symmetry

1. Given an equation,if an equivalent equation is obtained when y is replaced with −y, then the graph of the equation is symmetric with respect to the

x−axis.

Ex. x=y2

x = (−y)2

x = y2

∴symmetric wrt the x−axis

−1 1 2 3 4

Tests for Symmetry

1. Given an equation,if an equivalent equation is obtained when y is replaced with −y, then the graph of the equation is symmetric with respect to the

x−axis.

Ex. x=y2

x = (−y)2

x = y2

∴symmetric wrt the x−axis

−1 1 2 3 4

Tests for Symmetry

1. Given an equation,if an equivalent equation is obtained when y is replaced with −y, then the graph of the equation is symmetric with respect to the

x−axis.

Ex. x=y2

x = (−y)2

x = y2

∴symmetric wrt the x−axis

−1 1 2 3 4

Tests for Symmetry

1. Given an equation,if an equivalent equation is obtained when y is replaced with −y, then the graph of the equation is symmetric with respect to the

x−axis.

Ex. x=y2

x = (−y)2

x = y2

∴symmetric wrt the x−axis

−1 1 2 3 4

Tests for Symmetry

2. Given an equation,if an equivalent equation is obtained when x is replaced with −x, then the graph of the equation is symmetric with respect to the

y−axis.

Ex. y = 2x4

y= 2(−x)4

y= 2x4

∴symmetric wrt the y−axis −2−1 −1 21

Tests for Symmetry

2. Given an equation,if an equivalent equation is obtained when x is replaced with −x, then the graph of the equation is symmetric with respect to the

y−axis.

Ex. y = 2x4

y= 2(−x)4

y= 2x4

∴symmetric wrt the y−axis −2−1 −1 21

Tests for Symmetry

2. Given an equation,if an equivalent equation is obtained when x is replaced with −x, then the graph of the equation is symmetric with respect to the

y−axis.

Ex. y = 2x4

y= 2(−x)4

y= 2x4

∴symmetric wrt the y−axis −2−1 −1 21

Tests for Symmetry

2. Given an equation,if an equivalent equation is obtained when x is replaced with −x, then the graph of the equation is symmetric with respect to the

y−axis.

Ex. y = 2x4

y= 2(−x)4

y= 2x4

∴symmetric wrt the y−axis −2−1 −1 21

Tests for Symmetry

2. Given an equation,if an equivalent equation is obtained when x is replaced with −x, then the graph of the equation is symmetric with respect to the

y−axis.

Ex. y = 2x4

y= 2(−x)4

y= 2x4

∴symmetric wrt the y−axis −2−1 −1 21

Tests for Symmetry

3. Given an equation,if an equivalent equation is obtained when y with −y and x is replaced by

−x, then the graph of the equation is symmetric with respect to the origin.

Ex. y =x3

−y = (−x)3 −y =−x3

y =x3

∴symmetric wrt the origin

−2−1 1 2

Tests for Symmetry

3. Given an equation,if an equivalent equation is obtained when y with −y and x is replaced by

−x, then the graph of the equation is symmetric with respect to the origin.

Ex. y =x3

−y = (−x)3 −y =−x3

y =x3

∴symmetric wrt the origin

−2−1 1 2

Tests for Symmetry

3. Given an equation,if an equivalent equation is obtained when y with −y and x is replaced by

−x, then the graph of the equation is symmetric with respect to the origin.

Ex. y =x3

−y= (−x)3

−y =−x3 y =x3

∴symmetric wrt the origin

−2−1 1 2

Tests for Symmetry

3. Given an equation,if an equivalent equation is obtained when y with −y and x is replaced by

−x, then the graph of the equation is symmetric with respect to the origin.

Ex. y =x3

−y= (−x)3 −y=−x3

y =x3

∴symmetric wrt the origin

−2−1 1 2

Tests for Symmetry

3. Given an equation,if an equivalent equation is obtained when y with −y and x is replaced by

−x, then the graph of the equation is symmetric with respect to the origin.

Ex. y =x3

−y= (−x)3 −y=−x3

y=x3

∴symmetric wrt the origin

−2−1 1 2

Tests for Symmetry

3. Given an equation,if an equivalent equation is obtained when y with −y and x is replaced by

−x, then the graph of the equation is symmetric with respect to the origin.

Ex. y =x3

−y= (−x)3 −y=−x3

y=x3

∴symmetric wrt the origin

−2−1 1 2

Standard Equation of a Parabola

Definition

Let

a, b, c

∈

R

and

a

6

= 0. The graph of an

equation in

x

and

y

with the form

y

=

ax

+

bx

+

c

is called a (vertical)

parabola

.

Note: The equation

y

=

ax

+

bx

+

c, where

a, b, c

∈

R

and

a

6

= 0, is called the

standard form

Standard Equation of a Parabola

Definition

Let

a, b, c

∈

R

and

a

6

= 0. The graph of an

equation in

x

and

y

with the form

y

=

ax

+

bx

+

c

is called a (vertical)

parabola

.

Note: The equation

y

=

ax

+

bx

+

c, where

a, b, c

∈

R

and

a

6

= 0, is called the

standard form

Example

y=x2 _−2x

x y −2 8 −1 3 0 0 1 −1 2 0 3 3 4 8

−3−2−₋1₁ 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 9 10 0

(−2,8)

(−1,3)

(0,0)

(1,−1) (2,0)

Example

y=x2 _−2x

x y −2 8 −1 3 0 0 1 −1 2 0 3 3 4 8

−3−2−₋1₁ 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 9 10 0

(−2,8)

(−1,3)

(0,0)

(1,−1) (2,0)

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

1.

y

−

intercept :

c

2.

x

−

intercepts : real solutions to

ax

+

bx

+

c

= 0

• 2 real solutions ⇔ 2x−intercepts

• 1 real solution⇔ 1 x−intercept

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

1.

y

−

intercept :

c

2.

x

−

intercepts : real solutions to

ax

+

bx

+

c

= 0

• 2 real solutions ⇔ 2x−intercepts

• 1 real solution⇔ 1 x−intercept

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

1.

y

−

intercept :

c

2.

x

−

intercepts : real solutions to

ax

+

bx

+

c

= 0

• 2 real solutions ⇔2 x−intercepts

• 1 real solution⇔ 1 x−intercept

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

1.

y

−

intercept :

c

2.

x

−

intercepts : real solutions to

ax

+

bx

+

c

= 0

• 2 real solutions ⇔2 x−intercepts

• 1 real solution⇔ 1 x−intercept

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

1.

y

−

intercept :

c

2.

x

−

intercepts : real solutions to

ax

+

bx

+

c

= 0

• 2 real solutions ⇔2 x−intercepts

• 1 real solution⇔ 1 x−intercept

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

1.

y

−

intercept :

c

2.

x

−

intercepts : real solutions to

ax

+

bx

+

c

= 0

• 2 real solutions ⇔2 x−intercepts

• 1 real solution⇔ 1 x−intercept

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

3.

If

a >

0, the parabola opens upward.

If

a <

0, the parabola opens downward.

4.

vertex

(highest/lowest point of a parabola)

at the point with coordinates

−

b

2a

,

4ac

−

b

4a

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

3.

If

a >

0, the parabola opens upward.

If

a <

0, the parabola opens downward.

4.

vertex

(highest/lowest point of a parabola)

at the point with coordinates

−

b

2a

,

4ac

−

b

4a

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

3.

If

a >

0, the parabola opens upward.

If

a <

0, the parabola opens downward.

4.

vertex

(highest/lowest point of a parabola)

at the point with coordinates

−

b

2a

,

4ac

−

b

4a

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

3.

If

a >

0, the parabola opens upward.

If

a <

0, the parabola opens downward.

4.

vertex

(highest/lowest point of a parabola)

at the point with coordinates

−

b

2a

,

4ac

−

b

4a

Characteristics of a Parabola

Given

y

=

ax

+

bx

+

c,

3.

If

a >

0, the parabola opens upward.

If

a <

0, the parabola opens downward.

4.

vertex

(highest/lowest point of a parabola)

at the point with coordinates

−

b

2a

,

4ac

−

b

4a

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1)

,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 0

Parabolas

Ex. y=x2_−2x−3

y−int :

−3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1)

,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1)

,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0}

(x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1)

,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1)

,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1)

,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1)

,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1)

,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1) ,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4

0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1) ,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4

0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1) ,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4

0

Parabolas

Ex. y=x2_−2x−3

y−int : −3

x−int : x2_{−2x−3 = 0} (x+ 1)(x−3) = 0

x=−1 or x= 3

vertex :

−(−2) 2(1) ,

4(1)(−3)−(−2)2 4

= (1,−4)

−3−2−1 1 2 3 4 5 6

−7 −6 −5 −4 −3 −2 −1 1 2 3 4

0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1)

,

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int :

−6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1)

,

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1)

,

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1)

,

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1)

,

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...

No x−ints.

vertex :

−4 2(−1)

,

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1)

,

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1)

,

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1),

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1),

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1),

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Parabolas

Ex. y=−x2_{+ 4x−6}

y−int : −6

x−int : −x2+ 4x−6 = 0

x= −4±

p

42_{−4(−1)(−6)} 2(−1)

imaginary...No x−ints.

vertex :

−4 2(−1),

4(−1)(−6)−(4)2 4(−1)

= (2,−2)

−1 1 2 3 4 5

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 0

Circles

Definition

Acircle is a set of points in a plane equidistant from a fixed point called its center. The common distance is called the

radiusof the circle.

−2 −1 1 2 3 4

−1 1 2 3 4

Circles

Definition

Acircle is a set of points in a plane equidistant from a fixed point called its center. The common distance is called the

radiusof the circle.

−2 −1 1 2 3 4

−1 1 2 3 4

circle C

centered at (h, k) radius r

(h, k, r ∈_R, r >0)

−1 1 2 3 4 −2

−1 1 2 3 4

0

(h,k) r

If (x, y)∈C then p(x−h)2_{+ (y−k)}2 _{=r. (by distance} formula)

Theorem

Let h, k, r ∈_R and r >0.

An equation for the circle centered at (h, k) and with radius

r is

circle C

centered at (h, k) radius r

(h, k, r ∈_R, r >0)

−1 1 2 3 4 −2

−1 1 2 3 4

0

(h,k) r

If (x, y)∈C

then p(x−h)2_{+ (y−k)}2 _{=r. (by distance} formula)

Theorem

Let h, k, r ∈_R and r >0.

An equation for the circle centered at (h, k) and with radius

r is

circle C

centered at (h, k) radius r

(h, k, r ∈_R, r >0)

−1 1 2 3 4 −2

−1 1 2 3 4

0

(h,k) r

If (x, y)∈C then p(x−h)2_{+ (y−k)}2 _{=r. (by distance} formula)

Theorem

Let h, k, r ∈_R and r >0.

An equation for the circle centered at (h, k) and with radius

r is

circle C

centered at (h, k) radius r

(h, k, r ∈_R, r >0)

−1 1 2 3 4 −2

−1 1 2 3 4

0

(h,k) r

If (x, y)∈C then p(x−h)2_{+ (y−k)}2 _{=r. (by distance} formula)

Theorem

Let h, k, r ∈_R and r >0.

An equation for the circle centered at (h, k) and with radius

r is

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2 , radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂)

(−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂)

(−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

(−1 2,−

Example.

x+ 1 2

2

+

y−3 2 2 = 4 center: −1 2, 3 2

, radius= 2

−3 −2 −1 1 2

−1 1 2 3 4 0

(−1₂,3₂) (−5₂,3₂)

(−1₂,7₂)

(3₂,3₂)

(−1 2,−

The Unit Circle

The

unit circle

is the circle centered at (0

,

0)

and has radius 1

−2 −1 1 2

−2 −1 1 2

0

r

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

(x

−

2)

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

(x

−

2)

+ (y

+ 3)

Example: Find an equation for the circle with

center at

C

(2,

−

3) and passing through

P

(

−

1,

1).

r

=

|

P C

|

=

p

[2

−

(

−

1)]

+ [

−

3

−

1]

=

p

3

+ (

−

4)

=

√

9 + 16

= 5

Thus the equation is

Forms of the Equation of a Circle

Center-Radius Form

The center-radius form (also called standard form) of an equation of a circle is

(x−h)2+ (y−k)2 =r2.

By expanding the center-radius form..

General Form

An equation of a circle can be written in the general form

x2+y2+Dx+Ey+F = 0

Forms of the Equation of a Circle

Center-Radius Form

The center-radius form (also called standard form) of an equation of a circle is

(x−h)2+ (y−k)2 =r2.

By expanding the center-radius form..

General Form

An equation of a circle can be written in the general form

x2+y2+Dx+Ey+F = 0

Not every equation of the form (x−h)2+ (y−k)2 =d has a circle for its graph.

Remark

The graph of (x−h)2_{+ (y−k)}2 _=d:

• if d >0,

is a circle with center C(h, k) and radius√d

.

• if d= 0,

is the point P(h, k) (degenerate circle)

.

• if d <0,

is the empty set (imaginary circle)

Not every equation of the form (x−h)2+ (y−k)2 =d has a circle for its graph.

Remark

The graph of (x−h)2_{+ (y−k)}2 _=d:

• if d >0,

is a circle with center C(h, k) and radius√d

.

• if d= 0,

is the point P(h, k) (degenerate circle)

.

• if d <0,

is the empty set (imaginary circle)

Not every equation of the form (x−h)2+ (y−k)2 =d has a circle for its graph.

Remark

The graph of (x−h)2_{+ (y−k)}2 _=d:

• if d >0, is a circle with center C(h, k) and radius√d.

• if d= 0,

is the point P(h, k) (degenerate circle)

.

• if d <0,

is the empty set (imaginary circle)

Not every equation of the form (x−h)2+ (y−k)2 =d has a circle for its graph.

Remark

The graph of (x−h)2_{+ (y−k)}2 _=d:

• if d >0, is a circle with center C(h, k) and radius√d.

• if d= 0, is the point P(h, k) (degenerate circle).

• if d <0,

is the empty set (imaginary circle)

Not every equation of the form (x−h)2+ (y−k)2 =d has a circle for its graph.

Remark

The graph of (x−h)2_{+ (y−k)}2 _=d:

• if d >0, is a circle with center C(h, k) and radius√d.

• if d= 0, is the point P(h, k) (degenerate circle).

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(x

+

Dx

+

) +

(

y

+

Ey

+

) =

−

F

+

+

x

+

2

+

y

+

2

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2). • if D2_+E2_{−4F < 0, is}

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(

x

+

Dx

+

)

+

(

y

+

Ey

+

)

=

−

F

+

+

x

+

2

+

y

+

2

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2). • if D2_+E2_{−4F < 0, is}

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(

x

+

Dx

+

)

+

(

y

+

Ey

+

)

=

−

F

+

+

x

+

2

+

y

+

2

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2). • if D2_+E2_{−4F < 0, is}

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(

x

+

Dx

+

)

+

(

y

+

Ey

+

)

=

−

F

+

+

x

+

2

+

y

+

2

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2). • if D2_+E2_{−4F < 0, is}

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(

x

+

Dx

+

)

+

(

y

+

Ey

+

)

=

−

F

+

+

x

+

+

y

+

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2). • if D2_+E2_{−4F < 0, is}

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(x

+

Dx

+

) + (y

+

Ey

+

) =

−

F

+

+

x

+

2

+

y

+

2

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2). • if D2_+E2_{−4F < 0, is}

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(x

+

Dx

+

) + (y

+

Ey

+

) =

−

F

+

+

x

+

2

+

y

+

2

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2). • if D2_+E2_{−4F < 0, is}

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(x

+

Dx

+

) + (y

+

Ey

+

) =

−

F

+

+

x

+

2

+

y

+

2

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2).

• if D2_+E2_{−4F < 0, is}

Consider

x

+

y

+

Dx

+

Ey

+

F

= 0, where

D, E, F

∈

R

.

(x

+

Dx

+

) + (y

+

Ey

+

) =

−

F

+

+

x

+

2

+

y

+

2

=

The graph of the equation

x

+

y

+

Dx

+

Ey

+

F

= 0

• if D2_+E2_{−4F > 0, is a circle centered at (−}D

2,−

E

2) with radius

√

D2_+E2_−4F

2 .

• if D2_+E2_{−4F = 0, is the point P(−}D

2,−

E

2). • if D2_+E2_{−4F < 0, is}

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?}

Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F =

(−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2

+ (6)2−4(36) = 100 + 36−144 = −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2

−4(36) = 100 + 36−144 = −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+36= 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2₋₄₍₃₆₎

= 100 + 36−144 = −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100}

+ 36−144 = −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36}

−144 = −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144}

= −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0.

Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x

+ 9)

+

(

y2+ 10y

+ 25)

=−16

+ 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x+ 9

)

+

(

y2+ 10y

+ 25)

=−16 + 9

+ 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(

x2−6x+ 9

)

+

(

y2+ 10y+ 25

)

=−16 + 9 + 25

(x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(x2−6x+ 9) + (y2+ 10y+ 25) =−16 + 9 + 25 (x−3)2+ (y+ 5)2 = 18

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(x2−6x+ 9) + (y2+ 10y+ 25) =−16 + 9 + 25 (x−3)2+ (y+ 5)2 = 18

Thus, the center isC(3,−5)

Example:

What is the graph ofx2_+y2_{−10x+ 6y+ 36 = 0?} Solution:

D2+E2−4F = (−10)2_{+ (6)}2_{−4(36) = 100 + 36−144 =} −8<0. Hence, the graph of the equation is empty.

Example: Find the center and the radius of the circle with equationx2_+y2_{−6x+ 10y+ 16 = 0.}

Solution:

(x2−6x+ 9) + (y2+ 10y+ 25) =−16 + 9 + 25 (x−3)2+ (y+ 5)2 = 18

Tangent Lines

Tangent Lines

Determine the slope-intercept form of the equation of the line that is tangent to the circle with equation

x2 _+y2 _{−2x+ 4y−12 = 0 at P(0,2).}

Solution: Let ` be the line tangent to the circle at P(0,2):

(

x2−2x

+ 1)

+

(

y2 + 4y

+ 4)

= 12

+ 1 + 4 (x−1)2+ (y+ 2)2 = 17

` is perpendicular to line segment connecting the center

C(1,−2) to P(0,2):

Sincem_CP = 2−(−2) 0−1 =−4

,

we have m_` = 1 4

` has equation y−2 = 1

4(x−0) ⇔y=

Determine the slope-intercept form of the equation of the line that is tangent to the circle with equation

x2 _+y2 _{−2x+ 4y−12 = 0 at P(0,2).}

Solution: Let ` be the line tangent to the circle at P(0,2):

(

x2−2x

+ 1)

+

(

y2 + 4y

+ 4)

= 12

+ 1 + 4 (x−1)2+ (y+ 2)2 = 17

` is perpendicular to line segment connecting the center

C(1,−2) to P(0,2):

Sincem_CP = 2−(−2) 0−1 =−4

,

we have m_` = 1 4

` has equation y−2 = 1

4(x−0) ⇔y=

Determine the slope-intercept form of the equation of the line that is tangent to thecircle with equation

x2 _+y2 _{−2x+ 4y−12 = 0 atP(0,2).}

Solution: Let ` be the line tangent to the circle at P(0,2):

(

x2−2x

+ 1)

+

(

y2+ 4y

+ 4)