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Pre-Calc Equations of Circles Lesson 6.2

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

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Pre-Calc Equations of Circles Lesson 6.2

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  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 - √﴾(x1-x2)2 + (y1-y2)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 = r2

  2. So if any equation can be expressed in this ‘parent’ form (x-h)2 + (y-k)2 = r2 we can identify this as a circle with a center of (h,k) and will have a radius of √r2 = r If the center of the circle is (0,0), then the equation is simply: x2 + y2 = r2 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 = r2 ???  ??? (Duh!) Therefore Center  (h,k)  (3,-7) and r2 = 19 so r = √19

  3. x2 + y2 – 6x + 4y – 12 = 0 • does this look like x2 + y2 = r2 • Noooooooooooooooooooo!!!!!! • How about (x - h)2 + (y - k)2 = r2 ????? • Noooooooooooooooooooo!!!!!! • Not right now, but we can make this equation look like • this form---   (x - h)2 + (y - k) = r2 ????? • --- 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 x2 + y2 = 25 intersect. First of all the situation looks like this: Algebraically the process looks like this: y = 2x - 2 (1) x2 + y2 = 25 (2) (Hint: Use substitution)

  6. To find the intersection of a line and a circle algebraically: • Solve the linear equation for the easiest letter ‘x’ or ‘y’ • Substitute this expression for ‘y’ (or ‘x’) in the equation • of the circle. Then solve the resulting quadratic equation. • 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. • 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

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