Conic Sections: Circle

1. CONIC SECTION

2. CIRCLE

3. 1. Determine the radius and center of the equation
𝑥2
+𝑦2
− 8𝑥 + 4𝑦 − 16 = 0 .

4. In order to determine the radius and center of the circle, you
must write the equation in standard form
(𝑥 − ℎ)2
+(𝑦 − 𝑘)2
= 𝑟2
.

5. 𝑥2
− 8𝑥 + 𝑦2
+ 4𝑦 = 16
First step: Isolate the constant and group x’s and y’s
𝑥2
+𝑦2
− 8𝑥 + 4𝑦 − 16 = 0

6. 𝑥2
− 8𝑥 + 𝑦2
+ 4𝑦 = 16
Second step: Apply completing the square process, by adding a constant to both side
of the equation. This is done by dividing the value of b by 2 and square the result,
−8
2
2
= 16 𝑎𝑛𝑑
4
2
2
= 4.
𝑥2
− 8𝑥 + (_) + 𝑦2
+ 4𝑦 + (_) = 16 + (_)+(_)
𝑥2
− 8𝑥 + 16 + 𝑦2
+ 4𝑦 + 4 = 16 + 16 + 4

7. Third step: Write the perfect square trinomial in factored form and add the
constant at the right hand side of the equation.
(𝑥 − 4)(𝑥 − 4) + (𝑦 + 2)(𝑦 + 2) = 36
𝑥2
− 8𝑥 + 16 + 𝑦2
+ 4𝑦 + 4 = 16 + 16 + 4

8. (𝑥 − 4)(𝑥 − 4) + (𝑦 + 2)(𝑦 + 2) = 36
Fourth step: Express the final answer in square base form
(𝑥 − 4)2
+(𝑦 + 2)2
= 62

9. (𝑥 − 4)2
+(𝑦 + 2)2
= 62
Fifth step: Since the equation is already written in standard
form we can now determine the center and radius. Hence the
center (h,k) is (4, -2) and the radius is 6.
(𝑥 − ℎ)2
+(𝑦 − 𝑘)2
= 𝑟2

10. 2. Determine the equation of the circle given
the endpoints a diameter (-3, 6) and (3, -2).
In order to determine the equation of the circle, we must identify first
the center and radius then substitute it in the standard form.

11. Determine the radius using distance formula
𝑑 = (𝑥2 − 𝑥1)2+(𝑦2 − 𝑦1)2
𝑑 = (3 + 3)2+(−2 − 6)2
𝑑 = 36 + 64
𝑑 = 100
𝒅 = 𝟏𝟎 𝒖𝒏𝒊𝒕𝒔
r = 10/2 = 5 units
(-3, 6) and (3, -2).

12. (-3, 6) and (3, -2).Determine the center using midpoint formula
𝐶 =
𝑥1+𝑥2
2
,
𝑦1+𝑦2
2
𝐶 =
−3+3
2
,
−2+6
2
𝑪 = 𝟎, 𝟐

13. Now substitute these values 𝒓 = 𝟓 𝒖𝒏𝒊𝒕𝒔 and 𝑪 = 𝟎, 𝟐
in the standard form 𝑥 − ℎ 2 + 𝑦 − 𝑘 2 = 𝑟2
𝑥 − ℎ 2
+ 𝑦 − 𝑘 2
= 𝑟2

14. 𝑥 − ℎ 2
+ 𝑦 − 𝑘 2
= 𝑟2
First step: Substitute the values in the standard form
𝑥 − 0 2 + 𝑦 − 2 2 = 5 2

15. 𝑥 − 0 2 + 𝑦 − 2 2 = 5 2
Second step: Expand each expressions and combine like terms
𝑥2
+ 𝑦2
− 4𝑦 + 4 = 25

16. 𝑥2
+ 𝑦2
− 4𝑦 + 4 = 25
Third step: Express your answer in general form.
𝑥2
+ 𝑦2
− 4𝑦 − 21 = 0

17. 𝑃𝑅𝑂𝐵𝐿𝐸𝑀 𝑆𝑂𝐿𝑉𝐼𝑁𝐺
The danger zone is declared to be within a radius of 4
km from the site of explosion. Suppose the site of the
explosion is at the origin. Is a certain barangay located
at a place with coordinates (3, 3) within the danger
zone?

18. 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
To answer this problem, it is important to identify the given values,
make an illustration and identify the values being asked.

19. 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
In this case, the center is at the origin (0,0) with radius of 4
km. There are two ways to test if the coordinate (3, 3) is within
the danger zone or not. One is sketching the graph and the
other one is by substituting the this coordination in the
equation of the circle.

20. 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
• First Method: Graphing
Clearly on this graph, the point (3,3) is
outside the circle which means that it is
not within the danger zone.

21. 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
Second Method: Writing the Standard Form
Substitute these values 𝒓 = 𝟒 𝒌𝒎 and 𝑪 = 𝟎, 𝟎
in the standard form 𝑥 − ℎ 2 + 𝑦 − 𝑘 2 = 𝑟2

22. 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
𝑥 − 0 2
+ 𝑦 − 0 2
= 4 2
First step: Substitute the values in the standard form

23. 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
𝑥 − 0 2
+ 𝑦 − 0 2
= 4 2
𝑥2 + 𝑦2 = 16
Second step: Expand each expressions and combine like terms

24. 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
𝑥2 + 𝑦2 = 16
Third step: Substitute (3,3) in the equation and simplify the values.
(3)2
+(3)2
= 16

25. 𝑆𝑂𝐿𝑈𝑇𝐼𝑂𝑁
(3)2
+(3)2
= 16
Fourth step: make conclusion
18 > 16
Since the 18 is greater than 16, that means to say that (3, 3) is outside the danger zone.