1. Pre-Calc Equations of Circles Lesson 6.2 We can use the definition of a circle: A set of points each ‘equidistant’ from a fixed point—called the center, and the distance formula - √ ﴾(x 1 -x 2 ) 2 + (y 1 -y 2 ) 2 ﴿ to show how the ‘general form’ for the equation of all circles was derived: If the distance from ‘C’ (2,4) to a point ‘P’ (x,y) on the circle is 5  √﴾(x-2) 2 + (y-4) 2 ﴿ = 5 after squaring both sides  (x-2) 2 + (y-4) 2 = 25 If ‘C’ had been (h,k) and the distance from C to P = r then  (x-h) 2 + (y-k) 2 = r 2

2. So if any equation can be expressed in this ‘parent’ form (x-h) 2 + (y-k) 2 = r 2 we can identify this as a circle with a center of (h,k) and will have a radius of √r 2 = r If the center of the circle is (0,0), then the equation is simply: x 2 + y 2 = r 2 Example 1: Find the center and radius of each circle: a) (x - 3) 2 + (y + 7) 2 = 19 (Does this look like anything we have seen so far today?) (x - h) 2 + (y - k) 2 = r 2 ??? ??? (Duh!) Therefore Center  (h,k)  (3,-7) and r 2 = 19 so r = √19

3. b)x 2 + y 2 – 6x + 4y – 12 = 0 does this look like x 2 + y 2 = r 2 Noooooooooooooooooooo!!!!!! How about (x - h) 2 + (y - k) 2 = r 2 ????? Noooooooooooooooooooo!!!!!! Not right now, but we can make this equation look like this form---  (x - h) 2 + (y - k) = r 2 ????? --- can’t we ???? come on  (complete the square!) Go for it!

4. Example 2: a) Graph the equation in part 1(b) above with pencil and graph paper. b) Graph the equation in part 1(b) with your grapher.

5. Example 3: Find the coordinates of the point(s) where the line y = 2x - 2 and the circle x 2 + y 2 = 25 intersect. First of all the situation looks like this: Algebraically the process looks like this: y = 2x - 2 (1) x 2 + y 2 = 25 (2) (Hint: Use substitution)

6. To find the intersection of a line and a circle algebraically: 1.Solve the linear equation for the easiest letter ‘x’ or ‘y’ 2.Substitute this expression for ‘y’ (or ‘x’) in the equation of the circle. Then solve the resulting quadratic equation. 3.Substitute each real ‘x’-solution from step 2 in the linear equation to get the corresponding value of ‘y’ (or vice versa). Each point (x,y) is an intersection point. 4.You can check your result by substituting the coordinates of the intersection points in the two original equations. HW Pgs 222-223 # 1-25 odd