1. 6-2 Conic Sections: Circles
Geometric definition: A circle is formed by cutting a circular cone with a plane perpendicular to the symmetry axis of the cone. This intersection is a closed curve, and the intersection is parallel to the plane generating the circle of the cone.
Algebraic definition: A circle is the set of all points that are equally distant from a fixed point (the center).

2. Center-Radius Form 𝑥−ℎ 2 + 𝑦−𝑘 2 = 𝑟 2 ℎ,𝑘 =𝑐𝑒𝑛𝑡𝑒𝑟 𝑟=𝑟𝑎𝑑𝑖𝑢𝑠
𝑥−ℎ 𝑦−𝑘 2 = 𝑟 2
ℎ,𝑘 =𝑐𝑒𝑛𝑡𝑒𝑟
𝑟=𝑟𝑎𝑑𝑖𝑢𝑠
(h,k)
(x,y)
Example 1: Find the center and radius:
𝒙−𝟒 𝟐 + 𝒚+𝟕 𝟐 =𝟖𝟏 r
ℎ,𝑘 =(4,−7)
𝑟=9 x y

3. Example 2: Find the center and radius:
𝒙−𝟑 𝟐 + 𝒚+𝟏 𝟐 =𝟏𝟔
𝒄𝒆𝒏𝒕𝒆𝒓= 𝟑,−𝟏 𝒓=𝟒
Example 3: Find the center and radius:
𝒙+𝟓 𝟐 + 𝒚−𝟐 𝟐 =𝟏𝟓
𝒄𝒆𝒏𝒕𝒆𝒓= −𝟓,𝟐 𝒓= 𝟏𝟓
Example 4: Find the center and radius:
𝒙 𝟐 + 𝒚−𝟐 𝟐 =𝟗
𝒄𝒆𝒏𝒕𝒆𝒓= 𝟎,𝟐 𝒓=𝟑

4. Example 5: Write the equation of a circle centered at (2, 7) and having a radius of 5.
𝒙−𝟐 𝟐 + 𝒚+𝟕 𝟐 =𝟐𝟓
Example 6: Describe 𝒙−𝟐 𝟐 + 𝒚+𝟏 𝟐 =𝟎
𝑨 𝒑𝒐𝒊𝒏𝒕 𝒂𝒕 (𝟐, −𝟏)
Example 7: Describe 𝒙+𝟏 𝟐 + 𝒚−𝟑 𝟐 =−𝟏
𝑵𝒐 𝒈𝒓𝒂𝒑𝒉

5. Example 8: Rewrite in center-radius form by completing the square in x and y: 𝑥 2 −2𝑥+ 𝑦 2 −6𝑦=91 1 9
𝒙 𝟐 −𝟐𝒙+_____ + 𝒚 𝟐 −𝟔𝒚+_____ =𝟗+_____+_____ 9
− =1
𝑥− 𝑦−3 2 =19
− =9
𝒄𝒆𝒏𝒕𝒆𝒓= 𝟏,𝟑 𝒓= 𝟏𝟗
Example 9: Rewrite in center-radius form by completing the square in x and y: 𝑥 2 + 𝑦 2 −4𝑥+6𝑦+4=0 4 4 9
𝒙 𝟐 −𝟒𝒙+_____ + 𝒚 𝟐 +𝟔𝒚+_____ =−𝟒+_____+_____ 9
− =4
𝑥− 𝑦+3 2 =9 =9
𝒄𝒆𝒏𝒕𝒆𝒓= 𝟐,−𝟑 𝒓=𝟑

6. Example 10: Find an equation of the line tangent to circle 𝑥 2 + 𝑦 2 +4𝑥−46=0 at point P(3, 5).
Step 1: Write in center-radius form: (𝑥+2) 2 + (𝑦−0) 2 =46+4
(𝑥+2) 2 + (𝑦−0) 2 =50
Step 2: Check that P(3, 5) lies on the rim of the circle: (3+2) 2 + (5−0) 2 =50 ???
(5) 2 + (5) 2 =50
25+25=50
50=50
Step 3: Identify the center and radius: ℎ,𝑘 = −2,0 , 𝑟= 50 = =5 2
Step 4: Find slope of radius from center (2, 0) to P(3, 5):
𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 5−0 3−(−2) = 5 5 =1
Step 5: Write equation in point-slope form: 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1
𝑦−5=−1 𝑥−3

7. Step 1: Solve the linear equation for one variable: 𝑥=𝑦+3
Example 11: Find the intersection points between the circle 𝑥 2 + 𝑦 2 −10𝑥+4𝑦=−13 and the line 𝑥−𝑦=3:
Step 1: Solve the linear equation for one variable: 𝑥=𝑦+3
Step 2: Substitute into variable of circle equation: (𝑦+3) 2 + (𝑦) 2 −10 𝑦+3 +4𝑦=−13
𝑦 2 +6𝑦+9+ 𝑦 2 −10𝑦−30+4𝑦=−13
2 𝑦 2 −21=−13
2 𝑦 2 =8
𝑦 2 =4
𝑦=±2
Ax+BY=C
Stress that linear equations have exponents of 1 on both the x and the y.
With parabolas, either the x or the y has an exponent of 2.
With circles, both x and y terms are squared. Also with circles, the coefficients (or denominators) are the same for x2 and y2 (You will see in another lesson that that is not true for ellipses.)
Step 3: Substitute single-variable solutions into linear equation to solve for corresponding values: 𝑊ℎ𝑒𝑛 𝑦=2, 𝑥=5
𝑊ℎ𝑒𝑛 𝑦=−2, 𝑥=1
Solutions: (5, 2) and (1, 2)