This PowerPoint contains an introduction to conical sections: the conics formed from double-napped circular cone - the Parabola, Hyperbola, Circle, & Ellipse. It also contains the basic parts of Circle. Identifying the standard form of circle's radius and center. Graphing a circle from its standard form. Transforming General Equation of Circle to Standard Form and some of the special cases.

1. Chapter 1 – Analytic
Geometry
Lesson 1 - The Conic sections & Circles: An
Introduction
PRE-CALCULUS
PREPARED BY: SIR MYRRHTAIRE CASTILLO

2. Learning Objectives
At the end of the lesson, the student should be able to:
a. Illustrate the different types of conic sections: parabola,
ellipse, circle, hyperbola, and degenerate cases
b. Define a circle
c. Determine the equation of circle in standard form; and
d. Sketch a circle in a rectangular coordinate system.

3. *Interaction Part: Conics (if possible)

4. An Overview of Conic Sections
Conic sections are also
referred to as Conics.
Conics are the cross-
sectional view of a cone
(actually it is a double-
napped circular cone like
a cone with another cone on
its top joint through its apex
or tip).

5. When a plane cuts the cone
horizontally a circle is formed. As
shown in the image on the right the
lower portion depicts a circle. On
the other hand, when a plane cuts
the cone in a slant manner forming
a closed curve (bounded curve) an
ellipse is formed.
An Overview of Conic Sections

6. When a plane cuts the
cone in a slant manner
forming an open curve
(unbounded curve) a
parabola is formed.
An Overview of Conic Sections

7. When a plane cuts the cone
vertically forming two open
curves a hyperbola is
formed.
An Overview of Conic Sections

8. If you take a close look of the conics formed, its edges
(technically, it is called the cross-sectional area) formed would
look like this:
An Overview of Conic Sections
It's just like a
doodle but on the
next lessons and
chapter you will be
fascinated how
these doodles will
doodle your mind.
Anyways, let us
proceed with our
first conics -
CIRCLES.

9. Circles
1
9
7

10. A circle is the locus of all points in
the plane having the same fixed
positive distance, called the radius,
from a fixed point called center. We
are going to discover how the circle’s
standard form was derived but
before that let us take a short detour
on distance between two points.
Circles
Distance Between Two Points
To get the distance between two
points (𝑥1,𝑦1)and
(𝑥2, 𝑦2) remember this:
𝑑 = 𝑥2 − 𝑥1
2 + (𝑦2 − 𝑦1)2

11. Circles In the image, a circle with center
(h,k) which means the center NOT
on the origin and the radius r is the
segment from the center to a point
on the circle p. As we all know the
segment
𝑃𝐶 = 𝑟
Remember that the distance d
between two points is denoted by
𝑑 = 𝑥2 − 𝑥1
2 + (𝑦2 − 𝑦1)2

12. Circles
This time, our points are C(h,k)
and P(x,y) so interchanging (h,k)
to (𝑥1,𝑥2) and (x,y) to (𝑦1,𝑦2)
From
𝑃𝐶 = 𝑟
(𝑥 − ℎ)2+(𝑦 − 𝑘)2= 𝑟
to
To remove the root sign (radical)
square both sides.
(𝑥 − ℎ)2+(𝑦 − 𝑘)2= 𝑟2

13. Standard Equation of Circle
Center at (h,k)
Center at Origin (0,0)
(𝑥 − ℎ)2
+(𝑦 − 𝑘)2
= 𝑟2
𝑥2
+ 𝑦2
= 𝑟2

14. Example 1: Give the standard equation of
circle with the following given:
Center at Origin, radius 5.
Ans. 𝒙𝟐 + 𝒚𝟐 = 𝟐𝟓

15. Center (-2,6), radius 7
Example 2: Give the standard equation of
circle with the following given:
Ans. (𝒙 + 𝟐)𝟐+(𝒚 − 𝟔)𝟐= 𝟕

16. Example 3: Give the standard equation of
circle with the following given:
Center (-4,3) tangent to y-axis.
Ans. (𝒙 + 𝟒)𝟐
+(𝒚 − 𝟑)𝟐
= 𝟏𝟔

17. Example 4: Give the standard equation of
circle with the following given:
Center (-4,3) tangent to x-axis.
Ans. (𝒙 + 𝟒)𝟐
+(𝒚 − 𝟑)𝟐
= 𝟗

18. General Equation of Circle:Determining the center
and radius of circle
Say we have a circle with an equation of (𝑥 − 2)2+(𝑦 − 3)2= 4,
expanding this will give us
𝑥2
− 4𝑥 + 4 + 𝑦2
− 6𝑦 + 9 = 4
Combining the like terms and equating this equation will give us
𝑥2
+ 𝑦2
− 4𝑥 − 6𝑦 = 4 − 4 − 9
𝑥2 + 𝑦2 − 4𝑥 − 6𝑦 + 9 = 0
this is the general equation of circle
𝑤ℎ
𝑥2 + 𝑦2 + 𝐶𝑥 + 𝐷𝑦 + 𝐹 = 0
𝑤ℎ𝑒𝑟𝑒 𝐶 & 𝐷 ≠ 0

19. Example 5:Determine the center and radius of
the following Circles through the given
equation below.
𝑥2 + 𝑦2 − 5𝑥 = 7
Solution:
Regroup the terms
𝑥2 − 5𝑥 + 𝑦2 = 7
Completing the square (By completing the square, remember (
𝑏
2
)2 and since
our b=-5
𝑥2 − 5𝑥 +
25
4
+ 𝑦2 = 7 +
25
4

20. 𝑥2 − 5𝑥 +
25
4
+ 𝑦2 = 7 +
25
4
Factoring our x in (𝑥 − ℎ)2form
(𝑥 −
5
2
)2+𝑦2 =
53
4
Therefore, our Center is (
5
2
, 0)and radius is
53
2

21. Example 6:Determine the center and radius of
the following Circles through the given
equation below.
𝑥2 + 𝑦2 − 2𝑥 − 4𝑦 − 4 = 0
Solution:
Regroup the terms
𝑥2 − 2𝑥 + 𝑦2 − 4𝑦 = 4
Completing the square
(𝑥2 − 2𝑥) + (𝑦2−4𝑦) = 4
(𝑥2
−2𝑥 + 1) + (𝑦2
−4𝑦 + 4) = 4 + 1 + 4
Factoring the terms and simplifying the right hand side (it refers to the terms
on the right side of the equality sign)
(𝑥 − 1)2+(𝑦 − 2)2= 9
So, our center is (1,2)and radius is 3