graphs of equations conic sections-circles

1. Graphs of Equations

2. Graphs of Equations
There are two types of x&y formulas for graphing.

3. Graphs of Equations
There are two types of x&y formulas for graphing.
The simple type is where y can be expressed a function of x,
such as y = f(x) = –x2, so for an input x = a there is exactly one
output y = f(a).

4. Graphs of Equations
There are two types of x&y formulas for graphing.
The simple type is where y can be expressed a function of x,
such as y = f(x) = –x2, so for an input x = a there is exactly one
output y = f(a). To graph such a function, say y = f(x) = –x2,
we select values for the x, calculate the y’s, then plot the points
(x, y)’s to shape its graph.

5. Graphs of Equations
There are two types of x&y formulas for graphing.
The simple type is where y can be expressed a function of x,
such as y = f(x) = –x2, so for an input x = a there is exactly one
output y = f(a). To graph such a function, say y = f(x) = –x2,
we select values for the x, calculate the y’s, then plot the points
(x, y)’s to shape its graph.
y = –x2
x y
1 –1
2 –4
* *
* *

6. Graphs of Equations
There are two types of x&y formulas for graphing.
The simple type is where y can be expressed a function of x,
such as y = f(x) = –x2, so for an input x = a there is exactly one
output y = f(a). To graph such a function, say y = f(x) = –x2,
we select values for the x, calculate the y’s, then plot the points
(x, y)’s to shape its graph.
y = –x2
x y
1 –1
2 –4
* *
* *
The 2nd type of formulas is where it’s not possible to separate
and express the y as a single function in x.

7. Graphs of Equations
There are two types of x&y formulas for graphing.
The simple type is where y can be expressed a function of x,
such as y = f(x) = –x2, so for an input x = a there is exactly one
output y = f(a). To graph such a function, say y = f(x) = –x2,
we select values for the x, calculate the y’s, then plot the points
(x, y)’s to shape its graph.
y = –x2
x y
1 –1
2 –4
* *
* *
The 2nd type of formulas is where it’s not possible to separate
and express the y as a single function in x.
The simplest type of non–function–equations are the 2nd degree
equations such y2 + x2 = 1.

8. Graphs of Equations
There are two types of x&y formulas for graphing.
The simple type is where y can be expressed a function of x,
such as y = f(x) = –x2, so for an input x = a there is exactly one
output y = f(a). To graph such a function, say y = f(x) = –x2,
we select values for the x, calculate the y’s, then plot the points
(x, y)’s to shape its graph.
y = –x2
x y
1 –1
2 –4
* *
* *
The 2nd type of formulas is where it’s not possible to separate
and express the y as a single function in x.
The simplest type of non–function–equations are the 2nd degree
equations such y2 + x2 = 1. This leads to conic sections which
are the graphs of all 2nd degree equations (just as lines are
the graphs of the 1st degree equations).

9. Conic Sections
One way to study a solid is to slice it open.

10. Conic Sections
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.

11. Conic Sections
A right circular cone
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.

12. Conic Sections
A right circular cone and conic sections (wikipedia “Conic Sections”)
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.

13. Conic Sections
A Horizontal Section
A right circular cone and conic sections (wikipedia “Conic Sections”)
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.

14. Conic Sections
A Horizontal Section
A right circular cone and conic sections (wikipedia “Conic Sections”)
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.
Circles

15. Conic Sections
A Moderately
Tilted Section
A right circular cone and conic sections (wikipedia “Conic Sections”)
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.

16. Conic Sections
A Moderately
Tilted Section
A right circular cone and conic sections (wikipedia “Conic Sections”)
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.
Ellipses

17. Conic Sections
A Horizontal Section
A Moderately
Tilted Section
A right circular cone and conic sections (wikipedia “Conic Sections”)
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.
Circles and
ellipses are
enclosed.

18. Conic Sections
A right circular cone and conic sections (wikipedia “Conic Sections”)
A Parallel–Section
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.

19. Conic Sections
A right circular cone and conic sections (wikipedia “Conic Sections”)
A Parallel–Section
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.
Parabolas

20. Conic Sections
A right circular cone and conic sections (wikipedia “Conic Sections”)
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.
An Cut–away
Section

21. Conic Sections
A right circular cone and conic sections (wikipedia “Conic Sections”)
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.
An Cut–away
Section
Hyperbolas

22. Conic Sections
A right circular cone and conic sections (wikipedia “Conic Sections”)
An Cut–away
Section
One way to study a solid is to slice it open. The exposed area
of the sliced solid is called a cross sectional area.
Conic sections are the borders of the cross sectional areas of
a right circular cone as shown.
Parabolas and
hyperbolas are open.
A Horizontal Section
A Moderately
Tilted Section
Circles and
ellipses are
enclosed.
A Parallel–Section

23. Conic Sections
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.

24. Conic Sections
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections.

25. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.

26. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.
Graphs of
Ax2 + By2 + Cx + Dy = E,

27. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.
Graphs of
Ax2 + By2 + Cx + Dy = E,

28. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs. In some special cases their graphs degenerate into
lines or points, or nothing.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.

29. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs. In some special cases their graphs degenerate into
lines or points, or nothing.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.
x2 – y2 = 0
For example,
the graphs of:
x + y = 0x – y = 0

30. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs. In some special cases their graphs degenerate into
lines or points, or nothing.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.
x2 – y2 = 0
For example,
the graphs of:
x + y = 0x – y = 0
x2 + y2 = 0
(0,0)
(0,0) is the
only solution

31. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs. In some special cases their graphs degenerate into
lines or points, or nothing.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.
x2 – y2 = 0
For example,
the graphs of:
x + y = 0x – y = 0
x2 = –1x2 + y2 = 0
(0,0)
(0,0) is the
only solution
No solution
no graph

32. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs. In some special cases their graphs degenerate into
lines or points, or nothing.
We will match these 2nd degree equations with different conic
sections using the algebraic method "completing the square".
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.

33. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs. In some special cases their graphs degenerate into
lines or points, or nothing.
We will match these 2nd degree equations with different conic
sections using the algebraic method "completing the square".
“Completing the Square“ is THE main algebraic algorithm
for handling all 2nd degree formulas.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.

34. Conic Sections
(Most) Graphs of 2nd equations Ax2 + By2 + Cx + Dy = E,
are conic sections. The equations Ax2 + By2 + Cx + Dy = E
have conic sections that are parallel to the axes, i.e. not tilted,
as graphs. In some special cases their graphs degenerate into
lines or points, or nothing.
We will match these 2nd degree equations with different conic
sections using the algebraic method "completing the square".
“Completing the Square“ is THE main algebraic algorithm
for handling all 2nd degree formulas.
We need the Distance Formula D = Δx2 + Δy2 for the geometry.
Circles Ellipses Parabolas Hyperbolas
We summarize the four types of conic sections here.

35. Circles
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.

36. Circles
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
C

37. rr
Circles
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
C

38. rr
Circles
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
C

39. rr
The radius and the center completely determine the circle.
Circles
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
C

40. r
The radius and the center completely determine the circle.
Circles
Let (h, k) be the center of a
circle and r be the radius.
(h, k)
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
r
C

41. r
The radius and the center completely determine the circle.
Circles
(x, y)
Let (h, k) be the center of a
circle and r be the radius.
Suppose (x, y) is a point on
the circle, then the distance
between (x, y) and the center
is r.
(h, k)
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
r
C

42. r
The radius and the center completely determine the circle.
Circles
(x, y)
Let (h, k) be the center of a
circle and r be the radius.
Suppose (x, y) is a point on
the circle, then the distance
between (x, y) and the center
is r. Hence,
(h, k)
r =  (x – h)2 + (y – k)2
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
r
C

43. r
The radius and the center completely determine the circle.
Circles
(x, y)
Let (h, k) be the center of a
circle and r be the radius.
Suppose (x, y) is a point on
the circle, then the distance
between (x, y) and the center
is r. Hence,
(h, k)
r =  (x – h)2 + (y – k)2
or
r2 = (x – h)2 + (y – k)2
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
r
C

44. r
The radius and the center completely determine the circle.
Circles
(x, y)
Let (h, k) be the center of a
circle and r be the radius.
Suppose (x, y) is a point on
the circle, then the distance
between (x, y) and the center
is r. Hence,
(h, k)
r =  (x – h)2 + (y – k)2
or
r2 = (x – h)2 + (y – k)2
This is called the standard form of circles.
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
r
C

45. r
The radius and the center completely determine the circle.
Circles
(x, y)
Let (h, k) be the center of a
circle and r be the radius.
Suppose (x, y) is a point on
the circle, then the distance
between (x, y) and the center
is r. Hence,
(h, k)
r =  (x – h)2 + (y – k)2
or
r2 = (x – h)2 + (y – k)2
This is called the standard form of circles. Given an equation
of this form, we can easily identify the center and the radius.
A circle is the set of all the points that have equal distance r,
called the radius, to a fixed point C which is called the center.
r
C

46. r2 = (x – h)2 + (y – k)2
Circles

47. r2 = (x – h)2 + (y – k)2
must be “ – ”
Circles

48. r2 = (x – h)2 + (y – k)2
r is the radius must be “ – ”
Circles

49. r2 = (x – h)2 + (y – k)2
r is the radius must be “ – ”
(h, k) is the center
Circles

50. r2 = (x – h)2 + (y – k)2
r is the radius must be “ – ”
(h, k) is the center
Circles
Example B. Write the equation
of the circle as shown.
(–1, 3)

51. r2 = (x – h)2 + (y – k)2
r is the radius must be “ – ”
(h, k) is the center
Circles
Example B. Write the equation
of the circle as shown.
The center is (–1, 3)
and the radius is 5. (–1, 3)

52. r2 = (x – h)2 + (y – k)2
r is the radius must be “ – ”
(h, k) is the center
Circles
Example B. Write the equation
of the circle as shown.
The center is (–1, 3)
and the radius is 5.
Hence the equation is:
52 = (x – (–1))2 + (y – 3)2
(–1, 3)

53. r2 = (x – h)2 + (y – k)2
r is the radius must be “ – ”
(h, k) is the center
Circles
Example B. Write the equation
of the circle as shown.
The center is (–1, 3)
and the radius is 5.
Hence the equation is:
52 = (x – (–1))2 + (y – 3)2
or
25 = (x + 1)2 + (y – 3 )2
(–1, 3)

54. Example C. Identify the center and
the radius of 16 = (x – 3)2 + (y + 2)2.
Label the top, bottom, left and right
most points. Graph it.
Circles

55. Example C. Identify the center and
the radius of 16 = (x – 3)2 + (y + 2)2.
Label the top, bottom, left and right
most points. Graph it.
Put 16 = (x – 3)2 + (y + 2)2 into the
standard form:
42 = (x – 3)2 + (y – (–2))2
Circles

56. Example C. Identify the center and
the radius of 16 = (x – 3)2 + (y + 2)2.
Label the top, bottom, left and right
most points. Graph it.
Put 16 = (x – 3)2 + (y + 2)2 into the
standard form:
42 = (x – 3)2 + (y – (–2))2
Hence r = 4, center = (3, –2)
Circles

57. Example C. Identify the center and
the radius of 16 = (x – 3)2 + (y + 2)2.
Label the top, bottom, left and right
most points. Graph it.
Put 16 = (x – 3)2 + (y + 2)2 into the
standard form:
42 = (x – 3)2 + (y – (–2))2
Hence r = 4, center = (3, –2)
(3,–2)
Circles
r = 4

58. Example C. Identify the center and
the radius of 16 = (x – 3)2 + (y + 2)2.
Label the top, bottom, left and right
most points. Graph it.
Put 16 = (x – 3)2 + (y + 2)2 into the
standard form:
42 = (x – 3)2 + (y – (–2))2
Hence r = 4, center = (3, –2)
(3,–2)
Circles
r = 4

59. Example C. Identify the center and
the radius of 16 = (x – 3)2 + (y + 2)2.
Label the top, bottom, left and right
most points. Graph it.
Put 16 = (x – 3)2 + (y + 2)2 into the
standard form:
42 = (x – 3)2 + (y – (–2))2
Hence r = 4, center = (3, –2)
(3,–2)
Circles
When equations are not in the standard form, we have to
rearrange them into the standard form. We do this by
"completing the square".
r = 4

60. Example C. Identify the center and
the radius of 16 = (x – 3)2 + (y + 2)2.
Label the top, bottom, left and right
most points. Graph it.
Put 16 = (x – 3)2 + (y + 2)2 into the
standard form:
42 = (x – 3)2 + (y – (–2))2
Hence r = 4, center = (3, –2)
(3,–2)
Circles
When equations are not in the standard form, we have to
rearrange them into the standard form. We do this by
"completing the square".
To complete the square means to add a number to an
expression so the sum is a perfect square.
r = 4

61. Example C. Identify the center and
the radius of 16 = (x – 3)2 + (y + 2)2.
Label the top, bottom, left and right
most points. Graph it.
Put 16 = (x – 3)2 + (y + 2)2 into the
standard form:
42 = (x – 3)2 + (y – (–2))2
Hence r = 4, center = (3, –2)
(3,–2)
Circles
When equations are not in the standard form, we have to
rearrange them into the standard form. We do this by
"completing the square".
To complete the square means to add a number to an
expression so the sum is a perfect square. This procedure is
the main technique in dealing with 2nd degree equations.
r = 4

62. (Completing the Square)
Circles

63. (Completing the Square)
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

64. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square,
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

65. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square,
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

66. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square,
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

67. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square, i.e. x2 + bx + (b/2)2
is the perfect square (x + b/2)2.
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

68. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square, i.e. x2 + bx + (b/2)2
is the perfect square (x + b/2)2.
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2

69. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square, i.e. x2 + bx + (b/2)2
is the perfect square (x + b/2)2.
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2

70. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square, i.e. x2 + bx + (b/2)2
is the perfect square (x + b/2)2.
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36

71. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square, i.e. x2 + bx + (b/2)2
is the perfect square (x + b/2)2.
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2

72. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square, i.e. x2 + bx + (b/2)2
is the perfect square (x + b/2)2.
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2
The following are the steps in putting a 2nd degree equation
into the standard form.

73. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square, i.e. x2 + bx + (b/2)2
is the perfect square (x + b/2)2.
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2
The following are the steps in putting a 2nd degree equation
into the standard form.
1. Group the x2 and the x–terms together, group the y2 and y
terms together, and move the number term the the other
side of the equation.

74. (Completing the Square)
If we are given x2 + bx, then adding (b/2)2 to the expression
makes the expression a perfect square, i.e. x2 + bx + (b/2)2
is the perfect square (x + b/2)2.
Circles
Example D. Fill in the blank to make a perfect square.
a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2
The following are the steps in putting a 2nd degree equation
into the standard form.
1. Group the x2 and the x–terms together, group the y2 and y
terms together, and move the number term the the other
side of the equation.
2. Complete the square for the x–terms and for the y–terms.
Make sure you add the necessary numbers to both sides.

75. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
Circles

76. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
Circles

77. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36
Circles

78. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36 ;complete squares
x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36
Circles

79. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36 ;complete squares
x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36
Circles

80. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36 ;complete squares
x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36
( x – 3 )2 + (y + 6)2 = 9
Circles

81. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36 ;complete squares
x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36
( x – 3 )2 + (y + 6)2 = 9
( x – 3 )2 + (y + 6)2 = 32
Circles

82. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36 ;complete squares
x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36
( x – 3 )2 + (y + 6)2 = 9
( x – 3 )2 + (y + 6)2 = 32
Hence the center is (3 , –6),
and radius is 3.
Circles

83. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36 ;complete squares
x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36
( x – 3 )2 + (y + 6)2 = 9
( x – 3 )2 + (y + 6)2 = 32
Hence the center is (3 , –6),
and radius is 3.
Circles

84. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36 ;complete squares
x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36
( x – 3 )2 + (y + 6)2 = 9
( x – 3 )2 + (y + 6)2 = 32
Hence the center is (3 , –6),
and radius is 3.
Circles

85. Example E. Use completing the square to find the center
and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom,
left and right most points. Graph it.
We use completing the square to put the equation into the
standard form:
x2 – 6x + + y2 + 12y + = –36 ;complete squares
x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36
( x – 3 )2 + (y + 6)2 = 9
( x – 3 )2 + (y + 6)2 = 32
Hence the center is (3 , –6),
and radius is 3.
Circles
(3 ,–9)
(3 , –3)
(0 ,–6) (6 ,–6)

86. Conic Sections

87. Circles
B. Find the radius and the center of each circle.
Draw and label the four cardinal points.
1 = x 2 + y 2
1. 4 = x 2 + y 2
2.
1 = (x – 1)2 + (y – 3 )24. 4 = (x + 1)2 + (y + 5)25.
1/9 = x2 + y2
3.
1/4 = (x – 2)2 + (y + 3/2)26. 16 = (x + 5)2 + (y – 3)27.
25 = (x + 5)2 + (y – 4)28. 9 = (x + 1)2 + (y + 3)2
9.
1/16 = (x – 4)2 + (y – 6)210. 9/16 = (x + 7)2 + (y – 8)2
11.
1/100 = (x + 1.2)2 + (y – 4.1)212. 9/49 = (x + 9)2 + (y + 8)2
13.
0.01 = (x – 1.5 )2 + (y + 3.5)214.
0.49 = (x + 3.5)2 + (y + 0.5)2
15.

88. 2.
C. Complete the square to find the center and radius of each
of the following circles. Draw and label the four cardinal points.
1.
4.3.
6.5.
8.7.
13.
14.
Approximate the radius In the following problems.
x2 + y2 – 2y = 35
x2 – 8x + y2 + 12y = 92
x2 – 4y + y2 + 6x = –4
y2 – 8x + x2 + 2y = 8
x2 + 18y + y2 – 8x = 3
2x2 – 8x + 2y2 +12y = 1
x2 – 6x + y2 + 12y = –9
x2 + y2 + 8y = –7
x2 + 4x + y2 = 96
Circles
x2 + 6y + y2 – 16x = –9
10.9. x2 – x + y2 + 3y = 3/2 x2 + 3y + y2 – 5x = 1/2
12.
11.
x2 – 0.4x + y2 + 0.2y = 0.3
x2 + 0.8y + y2 – 1.1x = –0.04

89. The Mid-Point Formula
The mid-point m between two numbers a and b is the
average of them, that is m = .a + b
2
For example, the mid-point
a b(a+b)/2
mid-pt.
In the x&y coordinate the mid-point of (x1, y1) and (x2, y2) is
x1 + x2
2
,( y1 + y2
2
)
(x1, y1)
(x2, y2)
x1
y1
y2
x2
(x1 + x2)/2
(y1 + y2)/2
of 2 and 4 is (2 + 4)/2 = 3.
D. Given the two end points of
the diameter of a circle, use the
midpoint formula and the distance
formula to find the center, the
radius. Then find the equation of
each circle.
1. (3,1) and (1,3). 2. (–2,4) and (0,–3).
4. (6, –3) and (0,–3).3. (5,–4) and (–2,–1).

90. (Answers to odd problems) Ex. A
1. x2 + y2 = 25 5. (x – 1)2 + y2 = 93. x2 + y2 = 5
7. (x – 1)2 + (y + 1)2 = 81 9. (x + 3)2 + (y + 2)2 = 16
Circles
(0,5)
(0,-5)
(5,0)(-5,0)
(0,√5)
(0,-√5)
(√5,0)(-√5,0)
(1,3)
(1,-3)
(4,0)(-2,0)
(1,8)
(1,-10)
(10,0)(-8,0)
(-3,2)
(-3,-6)
(1,-2)(-2,-2)

91. Exercise B.
1. C = (0,0), r = 1. 5. C = (-1,-5), r = 2.3. C = (0,0), r = 1/3.
7. C = (-5,3), r = 4. 9. C = (-1,-3), r = 3. 11. C = (-7,8), r = ¾.
Circles
(0,1)
(0,-1)
(1,0)(-1,0)
(0,1/3)
(0,-1/3)
(1/3,0)(-1/3,0)
(-1,-3)
(-1,-7)
(1,-5)(-3,-5)
(-5,7)
(-5,-1)
(-1,3)(-9,3)
(-1,0)
(-1,-6)
(2,-3)(-4,-3)
(-7,8.75)
(-7,7.25)
(-6.25,8)
(-7.75,8)

92. 13. C = (-9,-8), r = 3/7. 15. C = (-3.5,-0.5), r = 0.7.
Exercise C.
5. C = (-3,2), r = 3.3. C = (0,-4), r = 3.1. C = (-2,0), r = 10.
Circles
(-9,-53/7)
(-9,-59/7)
(-60/7,-8)(-66/7,-8)
(-3.5,0.2)
(-3.5,-1.2)
(-2.8,-0.5)(-4.2,-0.5)
(-2,10)
(-2,-10)
(8,0)(-12,0)
(0,-1)
(0,-7)
(3,-4)(-3,-4)
(-3,5)
(-3,-1)
(0,2)(-6,2)

93. 7. C = (4,-6), r = 12. 9. C = (1/2,-3/2), r = 2.
13. C = (0.2,-0.1), r = 0.591611. C = (3, -6), r = 6.
Circles
(3,0)
(3,-12)
(8,-6)(-2,-6)
(1/2,1/2)
(1/2,-7/2)
(5/2,-3/2)(-3/2,-3/2)
(4,6)
(4,-18)
(16,-6)(-8,-6)

94. 1. C = (2, 2), r = √2.
Equation: (x – 2)2 + (y – 2)2 = 2
Exercise D.
3. C = (3/2, -5/2), r = (√58)/2.
Equation: (x – 3/2)2 + (y + 5/2)2 = 14.5
Circles