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2. CONIC
SECTIO
NS
Circles

3. Please send in chat your first
name and an adjective that
starts with the same letter as
your first name.
Example: Andrea Astig

4. Objectives
02
Illustrate and
define a circle
Graph a circle in a
rectangular
coordinate system
Determine the standard
form of the equation of
a circle
Practical
Exercises
01
04
03

6. - is the set of all point on a
plane that are equidistant
from a fixed point.
- This fixed point is called the
center,
- and the fixed distance is
called the radius
Circl
e
Center
Radius

7. Circle
Geometric
representation
Algebraic
representation

8. Standard Equation of a Circle
Line segment CP or
r = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
Given: C(h,k) as 𝑃1 and P(x,y) as
𝑃2
r = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
𝑟 = 𝑥 − ℎ 2 + 𝑦 − 𝑘 2
Squaring both the sides we
have:
(𝒙 − 𝒉)𝟐
+(𝒚 − 𝒌)𝟐
= 𝒓𝟐

9. Standard Equation of a Circle
Given: A(0,0) as 𝑃1 and B(x,y) as
𝑃2
r = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
𝑟 = 𝑥 − 0 2 + 𝑦 − 0 2
𝑟 = 𝑥2 + 𝑦2
Squaring both the sides we have:
𝒙𝟐
+ 𝒚𝟐
= 𝒓𝟐

10. Example 1 Solution
Problem: Find the equation
of the circle with the center
with center (3, -4) and radius
of 7 units.

11. We rewrite the standard form of the equation of the circle
in this form:
𝐴𝑥2
+ 𝐵𝑦2
+ 𝐶𝑥 + 𝐷𝑦 + 𝐸 = 0
or
We simply expand the standard equation.
General Equation of a circle

12. Problem: Find the equation
of the circle with the center
with center (3, -4) and radius
of 7 units.
Example 1 Solution
Standard Equation:
(𝒙 − 𝟑)𝟐
+(𝒚 + 𝟒)𝟐
= 𝟒𝟗

13. Example 1
Solution
Standard Equation:
Given:
ℎ = 3 ; 𝑘 = −4 ; 𝑟 = 7
(𝑥 − ℎ)2
+(𝑦 − 𝑘)2
= 𝑟2
(𝑥 − 3)2
+(𝑦 − (−4))2
= 72
(𝒙 − 𝟑)𝟐
+(𝒚 + 𝟒)𝟐
= 𝟒𝟗
General Equation:
𝐴𝑥2
+ 𝐵𝑦2
+ 𝐶𝑥 + 𝐷𝑦 + 𝐸 = 0
(𝑥 − 3)2
+(𝑦 + 4)2
= 49
𝑥2
− 6𝑥 + 9 + 𝑦2
+ 8𝑦 + 16 = 49
𝑥2
− 6𝑥 + 9 + 𝑦2
+ 8𝑦 + 16 − 49 = 0
𝒙𝟐
+ 𝒚𝟐
− 𝟔𝒙 + 𝟖𝒚 − 𝟐𝟒 = 𝟎
Problem: Find the equation
of the circle with the center
with center (3, -4) and radius
of 7 units.

14. Example 1:
Standard Equation:
(𝒙 − 𝟑)𝟐
+(𝒚 + 𝟒)𝟐
= 𝟒𝟗
General Equation:
𝒙𝟐
+ 𝒚𝟐
− 𝟔𝒙 + 𝟖𝒚 − 𝟐𝟒 = 𝟎

15. Example 1:
Standard Equation:
(𝒙 − 𝟑)𝟐
+(𝒚 + 𝟒)𝟐
= 𝟒𝟗
General Equation:
𝒙𝟐
+ 𝒚𝟐
− 𝟔𝒙 + 𝟖𝒚 − 𝟐𝟒 = 𝟎

16. Find the general equation
and standard equation of the
circle given the following
properties. Graph the
equation of the circle.
𝐶 −3, −5 ;
𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (1,7)
Example 2 Solution

17. Find the general equation
and standard equation of the
circle given the following
properties. Graph the
equation of the circle.
𝐶 −3, −5 ;
𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (1,7)
Example 2 Solution
𝒓 =?
Distance Formula
r = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
Given
𝐶(−3, −5) ; 𝑃(1,7)
Solution:
𝑟 = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
= 1 − (−3) 2 + 7 − (−5) 2
= 1 + 3 2 + 7 + 5 2
= 4 2 + 12 2
= 16 + 144
= 160
𝒓 = 𝟒 𝟏𝟎 or 12.65

18. Find the general equation
and standard equation of the
circle given the following
properties. Graph the
equation of the circle.
𝐶 −3, −5 ;
𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (1,7)
Example 2 Solution
Given
𝐶(−3, −5) ; 𝒓 = 𝟒 𝟏𝟎

19. Find the general equation
and standard equation of the
circle given the following
properties. Graph the
equation of the circle.
𝐶 −3, −5 ;
𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (1,7)
Example 2 Solution
Standard Equation:
(𝒙 + 𝟑)𝟐
+(𝒚 + 𝟓)𝟐
= 𝟏𝟔𝟎

20. Find the general equation
and standard equation of the
circle given the following
properties. Graph the
equation of the circle.
𝐶 −3, −5 ;
𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (1,7)
Example 2 Solution

21. Find the general equation
and standard equation of the
circle given the following
properties. Graph the
equation of the circle.
𝐶 −3, −5 ;
𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (1,7)
Example 2 Solution
Standard Equation:
Given:
ℎ = −3 ; 𝑘 = −5 ; 𝑟 = 4 10
(𝑥 − ℎ)2
+(𝑦 − 𝑘)2
= 𝑟2
(𝑥 − (−3))2
+(𝑦 − (−5))2
= 4 10
2
(𝒙 + 𝟑)𝟐
+(𝒚 + 𝟓)𝟐
= 𝟏𝟔𝟎
General Equation:
(𝑥 + 3)2
+(𝑦 + 5)2
= 160
𝑥2
+ 6𝑥 + 9 + 𝑦2
+ 10𝑦 + 25 = 160
𝑥2
+ 6𝑥 + 9 + 𝑦2
+ 10𝑦 + 25 − 160 = 0
𝒙𝟐
+ 𝒚𝟐
+ 𝟔𝒙 + 𝟏𝟎𝒚 − 𝟏𝟐𝟔 = 𝟎

22. Example 2:
Standard Equation:
(𝒙 + 𝟑)𝟐
+(𝒚 + 𝟓)𝟐
= 𝟏𝟔𝟎
General Equation:
𝒙𝟐
+ 𝒚𝟐
+ 𝟔𝒙 + 𝟏𝟎𝒚 − 𝟏𝟐𝟔 = 𝟎

23. Example 2:
Standard Equation:
(𝒙 + 𝟑)𝟐
+(𝒚 + 𝟓)𝟐
= 𝟏𝟔𝟎
General Equation:
𝒙𝟐
+ 𝒚𝟐
+ 𝟔𝒙 + 𝟏𝟎𝒚 − 𝟏𝟐𝟔 = 𝟎

24. Find the radius and the center of the circle defined by the equation.
2𝑥2 + 2𝑦2 + 8𝑥 − 6𝑦 + 2 = 0.
Example 3
Solution
Rewrite the expression in standard form in the form of…
(𝑥 − ℎ)2
+(𝑦 − 𝑘)2
= 𝑟2
Step 1: group the x terms
and y terms together and
transpose the constant to the
other side of the equation

25. Find the radius and the center of the circle defined by the equation.
2𝑥2 + 2𝑦2 + 8𝑥 − 6𝑦 + 2 = 0.
Example 3
Solution
Step 2: divide both
sides by two.
2𝑥2
+ 8𝑥 + 2𝑦2
− 6𝑦 = −1

26. Find the radius and the center of the circle defined by the equation.
2𝑥2 + 2𝑦2 + 8𝑥 − 6𝑦 + 2 = 0.
Example 3
Solution
Step 3: Complete
the square by
adding the square of
half of the middle
term on both sides
of the equation.
= −1
𝑥2 + 4𝑥 +𝑦2 − 3𝑦

27. Find the radius and the center of the circle defined by the equation.
2𝑥2 + 2𝑦2 + 8𝑥 − 6𝑦 + 2 = 0.
Example 3
Solution
Step 4: Express the perfect
square trinomial as the
square of a binomial.
𝑥2
+ 4𝑥 + 4 + 𝑦2
− 3𝑦 +
9
4
=
21
4

28. Find the radius and the center of the circle defined by the equation.
2𝑥2 + 2𝑦2 + 8𝑥 − 6𝑦 + 2 = 0.
Example 3
Solution
2𝑥2 + 8𝑥 + 2𝑦2 − 6𝑦 = −2
𝑥2
+ 4𝑥 + 4 + 𝑦2
− 3𝑦 +
9
4
=
21
4
𝒙 + 𝟐 𝟐 + 𝒚 −
𝟑
𝟐
𝟐
=
𝟐𝟏
𝟒
Thus,
Standard Equation
𝑥 + 2 2 + 𝑦 −
3
2
2
=
21
4
where;
ℎ = −2 ; 𝑘 =
3
2
and 𝑟 =
21
2

29. Questions

30. What are your major takeaways in
today’s discussion?

32. Please send in chat your first
name and an adjective that
starts with the same letter as
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37. Have great
week
ahead! ♥
-Ms. Andrea