The document discusses conic sections, which are curves formed by the intersection of a plane and a right circular cone. There are four types of conic sections: circles, ellipses, parabolas, and hyperbolas. Conic sections can be represented by second-degree equations in x and y, and the technique of completing the square is used to determine which equation corresponds to each type of conic section. The document also reviews the distance formula.

1. Conic Sections

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

3. 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.

4. 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.

5. 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.

6. 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.

7. 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.

8. 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.

9. 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.

10. 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
ellipsis are
enclosed.

11. 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.

12. 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.

13. 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

14. 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

15. 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
ellipsis are
enclosed.
A Parallel–Section

16. We summarize the four types of conics sections here.
Circles Ellipses
Parabolas Hyperbolas
Conic Sections

17. Conic Sections

18. Conic Sections
Conic sections are the cross sections of right circular cones.
There are four different types of curves:
• circles
• ellipses

19. Conic Sections
Conic sections are the cross sections of right circular cones.
There are four different types of curves:
• circles
• ellipses
• parabolas

20. Conic Sections
Conic sections are the cross sections of right circular cones.
There are four different types of curves:
• circles
• ellipses
• parabolas
• hyperbolas
Where as straight lines are the graphs of 1st degree equations
Ax + By = C, conic sections are the graphs of 2nd degree
equations in x and y.

21. Conic Sections
Conic sections are the cross sections of right circular cones.
There are four different types of curves:
• circles
• ellipses
• parabolas
• hyperbolas
Where as straight lines are the graphs of 1st degree equations
Ax + By = C, conic sections are the graphs of 2nd degree
equations in x and y. In particular, the conic sections that are
parallel to the axes (not tilted) have equations of the form
Ax2
+ By2
+ Cx + Dy = E, where A, B, C, D, and E are
numbers (not both A and B equal to 0).

22. Conic Sections
Conic sections are the cross sections of right circular cones.
There are four different types of curves:
• circles
• ellipses
• parabolas
• hyperbolas
Where as straight lines are the graphs of 1st degree equations
Ax + By = C, conic sections are the graphs of 2nd degree
equations in x and y. In particular, the conic sections that are
parallel to the axes (not tilted) have equations of the form
Ax2
+ By2
+ Cx + Dy = E, where A, B, C, D, and E are
numbers (not both A and B equal to 0). We are to match these
2nd degree equations with the different conic sections.

23. Conic Sections
Conic sections are the cross sections of right circular cones.
There are four different types of curves:
• circles
• ellipses
• parabolas
• hyperbolas
Where as straight lines are the graphs of 1st degree equations
Ax + By = C, conic sections are the graphs of 2nd degree
equations in x and y. In particular, the conic sections that are
parallel to the axes (not tilted) have equations of the form
Ax2
+ By2
+ Cx + Dy = E, where A, B, C, D, and E are
numbers (not both A and B equal to 0). We are to match these
2nd degree equations with the different conic sections.
The algebraic technique that enable us to sort out which
equation corresponds to which conic section is called
"completing the square".

24. Conic Sections
Conic sections are the cross sections of right circular cones.
There are four different types of curves:
• circles
• ellipses
• parabolas
• hyperbolas
Where as straight lines are the graphs of 1st degree equations
Ax + By = C, conic sections are the graphs of 2nd degree
equations in x and y. In particular, the conic sections that are
parallel to the axes (not tilted) have equations of the form
Ax2
+ By2
+ Cx + Dy = E, where A, B, C, D, and E are
numbers (not both A and B equal to 0). We are to match these
2nd degree equations with the different conic sections.
The algebraic technique that enable us to sort out which
equation corresponds to which conic section is called
"completing the square". We start with the Distance Formula.

25. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
Conic Sections

26. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
Conic Sections

27. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
Conic Sections

28. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
where
Conic Sections
Δy = the difference between the y's = y2 – y1

29. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
where
Conic Sections
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1

30. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
where
Example A. Find the
distance between (2, –1)
and (–2, 2).
Conic Sections
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1

31. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
where
Example A. Find the
distance between (2, –1)
and (–2, 2).
Δy = (–1) – (2) = –3
Δy=-3
Conic Sections
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1

32. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
where
Example A. Find the
distance between (2, –1)
and (–2, 2).
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
Δy=-3
Δx=4
Conic Sections
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1

33. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
where
Example A. Find the
distance between (2, –1)
and (–2, 2).
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
r = √(–3)2
+ 42
= √25 = 5
Δy=-3
Δx=4
r=5
Conic Sections
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1

34. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
where
Example A. Find the
distance between (2, –1)
and (–2, 2).
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
r = √(–3)2
+ 42
= √25 = 5
Δy=-3
Δx=4
r=5
Conic Sections
The geometric definition of all four types of conic sections are
distance relations between points.
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1

35. The Distance Formula:
Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane,
the distance r between P and Q is:
r = √(y2 – y1)2
+ (x2 – x1)2
= √Δy2
+ Δx2
where
Example A. Find the
distance between (2, –1)
and (–2, 2).
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
r = √(–3)2
+ 42
= √25 = 5
Δy=-3
Δx=4
r=5
Conic Sections
The geometric definition of all four types of conic sections are
distance relations between points. We start with the circles.
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1

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.

37. r
r
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. r
r
The radius and the center completely determine the circle.
Circles
center
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.

39. 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.

40. 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.

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. 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.

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
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.

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
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.

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. 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.

45. r2
= (x – h)2
+ (y – k)2
Circles

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

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

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

49. 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.

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.
The center is (–1, 3) and the
radius is 5. (–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.
Hence the equation is:
52
= (x – (–1))2
+ (y – 3)2
(–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
or
25 = (x + 1)2
+ (y – 3 )2
(–1, 3)

53. 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

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.
Put 16 = (x – 3)2
+ (y + 2)2
into the
standard form:
42
= (x – 3)2
+ (y – (–2))2
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
Hence r = 4, center = (3, –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)
(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
(3, 2)
(3,-6)

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
(3, 2)
(–1,-2) (7,-2)
(3,-6)

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
(3, 2)
(–1,-2) (7,-2)
(3,-6)

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)
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".
(3,-2)
(3, 2)
(–1,-2) (7,-2)
(3,-6)

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)
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.
(3,-2)
(3, 2)
(–1,-2) (7,-2)
(3,-6)

62. 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
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.
(3,-2)
(3, 2)
(–1,-2) (7,-2)
(3,-6)

63. (Completing the Square)
Circles

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

65. (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

66. (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

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

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

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

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 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
(3, –6)

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
(3, –6)
(3, –3)
(3, –9)
(6, –6)(0, –6)

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, –6)
(3, –3)
(3, –9)
(6, –6)(0, –6)

86. Conic Sections

87. Conic Sections

88. Conic Sections

89. Conic Sections
21. x2
+ y2
= 4
Use the completing the square method to write the equation
in the standard form of the circles. Where is the center?
What is the radius? Draw the circle and label the four cardinal
points on the circle.
22. x2
+ y2
= 16
23. x2
– 4x + y2
= 0 24. x2
+ y2
+ 2y = 24
25. x2
– 2x + y2
+ 4y = –4 26. x2
– 6x + y2
+ 4y = 3
27. x2
+ 12y + y2
– 8x = – 3 28. x2
+ y2
– 10x + 2y = –1
29. x2
– 18y + y2
– 8x = 3 30. x2
+ y2
– 14x + 2y = 14

90. Conic Sections

91. Conic Sections

92. Conic Sections
Answers