1. Day 129 – Equation of a circle centered at any point

2. Introduction
The key feature of a circle is that every point on the circle is the same distance from the center of the circle. In the previous lesson, we derived the equation of a circle centered at the origin. The equation was based on the center and radius of the circle. Also, we used the Pythagorean theorem when deriving the equation of a circle. In this lesson, we will also apply the Pythagorean theorem and learn how to complete the square in order to find the center and radius of a circle whose center is a general point when given the equation of that circle.

3. Vocabulary
1. Radius
The distance from the center of a circle to any point on that circle.
2. Center of a circle
A point inside a circle such that all points on the circle are equidistant from that point.
3. A perfect square(in expressions)
A quadratic expression which forms two identical factors when factorized.

4. The general equation of a circle
The Pythagorean theorem is vital when it comes to deriving the equation of a circle centered at any point, that is, the general equation of a circle. We will also apply the distance formula to derive this equation. Once we have derived the equation of the circle, we will learn how to complete the square in order to find the center and the radius of the circle when we are given the equation of that circle.

5. In the figure below, the circle is centered at 𝐶 𝑎, 𝑏 and the radius is 𝐶𝐾=𝑟.𝑥 𝑦
K 𝑥, 𝑦
C 𝑎, 𝑏
L 𝑥 , 𝑏 O 𝑟
𝑥−𝑎
𝑦−𝑏

6. Right ∆CKL is formed such that 𝐶𝐾=𝑟 is the hypotenuse
Right ∆CKL is formed such that 𝐶𝐾=𝑟 is the hypotenuse. Using the distance formula, we find out that: 𝐶𝐿=𝑥−𝑎 and 𝐾𝐿=𝑦−𝑏 If we use the Pythagorean theorem, we have: CL 2 + KL 2 = CK 2 but CL=𝑥−𝑎 and KL=𝑦−𝑏 Therefore, we have: 𝒙−𝒂 𝟐 + 𝒚−𝒃 𝟐 = 𝒓 𝟐

7. The equation 𝒙−𝒂 𝟐 + 𝒚−𝒃 𝟐 = 𝒓 𝟐 is the general equation of a circle centered at an arbitrary point 𝐶 𝑎,𝑏 and having radius 𝑟. This is the equation of a circle at any point. If the center of the circle is at the origin, 0, 0 the equation; 𝒙−𝒂 𝟐 + 𝒚−𝒃 𝟐 = 𝒓 𝟐 changes to 𝒙 𝟐 + 𝒚 𝟐 = 𝒓 𝟐 In general,
Equation
Center
Radius
𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2
𝑎, 𝑏 𝑟
𝑥+𝑎 𝑦−𝑏 2 = 𝑟 2
−𝑎, 𝑏
𝑥−𝑎 𝑦+𝑏 2 = 𝑟 2
𝑎, −𝑏
𝑥+𝑎 𝑦+𝑏 2 = 𝑟 2
−𝑎, −𝑏

8. Completing the square
Some quadratic equations can be easily solved because when we factorize the left hand side, we get two identical factors. This means that the left hand side is a perfect square. We can also get to distinct factors. For example, the quadratic equation; 𝑥 2 +6𝑥+9=0 When we factorize the left hand side, 𝑥 2 +6𝑥+9, we get to identical factors; 𝑥+3 and 𝑥+3 We therefore have; 𝑥+3 𝑥+3 =0⇒ 𝑥+3 2 =0

9. This, however, is not the case for all quadratic equations
This, however, is not the case for all quadratic equations. A number of quadratic equations have quadratic expressions that cannot factorized into their respective factors. We therefore need to apply the concept of completing the square to make the quadratic expressions in the quadratic equations factorable. Completing the square means making the quadratic expression a perfect square.

10. A quadratic expression is of the form 𝑎𝑥 2 +𝑏𝑥+𝑐
A quadratic expression is of the form 𝑎𝑥 2 +𝑏𝑥+𝑐. If 𝑎=1, the expression becomes 𝑥 2 +𝑏𝑥+𝑐. If we are given the part 𝑥 2 +𝑏𝑥, we can find the value of 𝑐 that makes the expression a perfect square. We use the relation: 𝑏 2 2 =𝑐

11. Example 1 Complete the square given the following expressions: (a) 𝑥 2 −8𝑥 (b) 𝑦 2 +12𝑦 Solution The missing term is 𝑐 hence we use the relation 𝑏 2 2 =𝑐 𝑥 2 −8𝑥+ −8 2 2 = 𝑥 2 −8𝑥+16

12. (b) 𝑦 2 +12𝑦 The missing term is 𝑐 hence we use the relation 𝑏 2 2 =𝑐 𝑦 2 +12𝑦+ 12 2 2 = 𝑦 2 +12𝑦+36

13. Example 2 The equation of a circle is 𝑥 2 + 𝑦 2 −2𝑥+8𝑦−1=0
Example 2 The equation of a circle is 𝑥 2 + 𝑦 2 −2𝑥+8𝑦−1=0. Determine the center and radius of the circle. Solution The equation of the circle is 𝑥 2 + 𝑦 2 −2𝑥+8𝑦−1=0 We first put the terms with 𝑥 together and the terms with 𝑦 together. The equation becomes: 𝑥 2 −2𝑥+ 𝑦 2 +8𝑦−1=0…(i) We then take the constant to the right hand side of the equation. 𝑥 2 −2𝑥+ 𝑦 2 +8𝑦=1…(ii)

14. We then complete the square of both the 𝑥 terms and 𝑦 terms on the left hand side of (ii). 𝑥 2 −2𝑥+ −2 2 2 and 𝑦 2 +8𝑦 𝒙 𝟐 −𝟐𝒙+ −𝟐 𝟐 𝟐 + 𝒚 𝟐 +𝟖𝒚+ 𝟖 𝟐 𝟐 =𝟏+ −𝟐 𝟐 𝟐 + 𝟖 𝟐 𝟐 …(𝐢𝐢𝐢) 𝑥−1 2 + 𝑦+4 2 =1+1+16=18

15. 𝑥−1 2 + 𝑦+4 2 =18…(iv) From (iv) , the center of the circle is 1, −4 and the radius is 18 .

16. homework
Find the center and radius of a circle with equation 𝑥 2 + 𝑦 2 +2𝑥−8𝑦+8=0.

17. Answers to homework
Center −1, 4 , radius 3 units

18. THE END