1. Circles Tools we need for circle problems:
Midpoint formula: This is used to find the point exactly between two points. We can use this to find the center of a circle.
Ex. Find midpt. of (4,3) and (-2,7)
Distance formula: This is used to find the distance between two points. We can use it to find the radius of a circle.
Ex. Find the distance between (1,2) and (-2,7)

2. Circles tools continued
Standard form of a circle: This form is used to write an equation of a circle when you know the center and the radius.
(x – h)2 + (y – k)2 = r2 where (h,k) is the center of the circle and r is the length of the radius.
Ex. Center is (3,-4) radius is (x – 3)2 + (y + 4)2 = 16
General form of a circle: This form is used to write an equation of a circle in a form that is polynomial in looks.
x2 + y2 +ax + by + c = 0
Completing the square: This is used to convert the equation of a circle in general form to standard form. Remember x2 + 6x = 0 can be completed by:
x2 + 6x + 9 = 9
(x + 3)2 = 9

3. Circles
Every point on a circle must be equal distance from a point called the center and the distance is called the radius.
The standard equation is x2 + y2 = r2 if the center is at the origin.
The equation for this circle would be x2 + y2 = 52 or
x2 + y2 = 25
A circle with a center at the origin and radius of 1 is written as x2 + y2 = 1 and this is called the UNIT circle.
Radius

4. Using these tools to solve problems.
Find the center and radius of the given circle and write the standard form of the equation:
Ex. C(1,2) Pt on circle (1,0) Practice: pts on the circle making a diameter (1,2), (4,2)
How can we find the center and radius
Center
Radius:
S.F (x – 5/2)2 + (y – 2)2 = 9/4
Radius =
S.F = (x – 1)2 + (y – 2)2 = 4

5. Writing both types of equations:
Given the center and the radius write the equation of a circle in standard and general form.
Ex. C(0,2) r = 2 practice: C(4,-3) r = 5
S.F. x2 + (y – 2)2 = 4
How can we change it to general form?
G.F. x2 + y2 – 4y + 4 = 4
x2 + y2 – 4y = 0
S.F. (x – 4)2 + (y + 3)2 = 25
G.F. x2 – 8x y2 + 6y + 9 = 25
x2 + y2 – 8x + 6y = 0

6. Find the center, the radius, any x or y intercepts, and graph.
Ex. 3(x + 1)2 + 3(y – 1)2 = 6
This equation is almost in standard form what do we need to do to get it into standard form.
(x + 1)2 + (y – 1)2 = 2
The center is (-1,1) and the radius is √2 Now how can we find the x and y intercepts hint we it with lines.
To find the x intercept set y = 0
(x + 1)2 + (0 – 1)2 = 2
(x + 1)2 + 1 = 2
(x + 1)2 = 1
x + 1 = ± 1
x = 0 x = -2
(0,0) (0,-2)
To find the y intercept set x = 0
(0 + 1)2 + (y – 1)2 = 2
1 + (y – 1)2 = 2
(y – 1)2 = 1
y – 1 = ± 1
y = 0 x = 2
(0,0) (2,0)

7. Find the center, the radius, any x or y intercepts, and graph.
Practice: x2 + y2 – x + 2y + 1 = 0
This equation is in general form what do we need to do to get it into standard form.
We need to complete the square for both the x’s and y’s
x2 – x + ¼ y2 + 2y + 1 = -1 + ¼ + 1
(x – ½ )2 + (y + 1)2 = ¼
Now we have it in standard form so we can find the center and radius
The center is ( ½ ,-1) and the radius is ½ Now how can we find the x and y intercepts hint we it with lines.
To find the x intercept set y = 0
(x - ½ )2 + (0 + 1)2 = ¼
(x - ½ )2 + 1 = ¼
(x - ½ )2 = - ¾
no real solutions
To find the y intercept set x = 0
(0 – ½ )2 + (y + 1)2 = ¼
¼ + (y + 1)2 = ¼
(y + 1)2 = 0
y + 1 = 0
y = -1
(0,-1)

8. Find the standard form of the equation with the given information
Ex. Center (4,-2) and tangent to the line x = 1
We have the center but need the radius. What does a tangent line do?
How can we use that to find the radius?
A tangent line touches a circle at one point and that point in this case has to be when x = 1 so all we need to know is how far is x = 1 to the center x value? 1 to 4 is 3 so the radius is 3
So the equation is (x – 4)2 + (y + 2)2 = 9
Practice: Center (2,3) and tangent to the x axis.
x axis is y = 0 so the radius is 0 to 3 which is 3
(x – 2)2 + (y – 3)2 = 9