# NOTES: Math 114: Circle Format, Completing the

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```Math 114 Notes : Circle Equation
The DISTANCE FORMULA is d  ( x2  x1 ) 2  ( y2  y1 ) 2
where the two points we want to find the distance between is ( x1 , y1 ) and ( x2 , y2 ) .
Suppose we change the variables in the distance formula so that we have
r  ( x  h) 2  ( y  k ) 2 .
r 2  ( x  h)2  ( y  k )2 . An example of this equation is
52  ( x  0)2  ( y  0) 2 .
Below is a table of values (points) for this equation and a graph.
Now, square both sides. We get
The graph is a CIRCLE with center (0, 0) and radius 5 (count the tic marks on the x-axis to see
that the radius is 5).
We started with the equation r 2  ( x  h)2  ( y  k )2 . The radius is r and the center is (h, k).
CIRCLE EQUATION: r 2  ( x  h)2  ( y  k )2 radius = r center = (h, k)
Find the center and radius of the following circles using the circle equation.
1. ( x  5) 2  ( y  6) 2  49
2. x 2  y 2  36 (h int : ( x  0) 2  ( y  0) 2  36)
3. ( x  2) 2  ( y  3) 2  1
For #1,
For #2,
For #3,
( x  5) 2  ( y  6) 2  49
( x  h)  ( y  k )  r
( x  0) 2  ( y  0) 2  36
2
2
2
( x  h)  ( y  k )  r
( x  2) 2  ( y  3) 2  1
2
( x  h)  ( y  k )  r
2
2
2
2
2
so h = 5, k = 6 and r 2  49 . Center is (5, 6); Radius is 7.
so h = 0, k = 0 and r 2  36 . Center is (0, 0); Radius is 6
so h = 2, k = - 3 and r 2  1 . Center = (2, -3); Radius = 1.
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