How To Determine Circle Equations

II.        LOOK FOR CIRCLE EQUATIONS ^^

         Now the reverse, how to find the equation of the circle if known its center and radius.

     Determine the equation of the circle below :

1.     Center (0,0) r = 3

2.     Center (-3, -1) r = 6
3.   Point (6, 8) is on a Circle A the center is (0.0). The circle equations?
            Answer 

  1. For the center (0,0) The circle equation  : x² + y² = r²     

o   So, the equation circle = x² + y² = 3

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2. For the center not (0,0) r The circle equation  : (x-a)² + (y-b)² =r²

o   So, the equation circle = (  x-(-3) )² + ( y – (-8) )² = 6²

o   The equation circle = (x+3)² + (y+8)² = 6²

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3. For the Center (0,0) and point (6.8) are in the circle, 
because passed on the circle. (6.8) as (x, y)
• Equation circle when center (0,0) = x² + y² = r²   (x,y) = (6,8)
Because radius unknown we seek it first:
x² + y² = r²
6² + 8² = r²
r² = 100
then the equation circle = x² + y² = 10²

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So now, how to determine Circle Equations if the circle passed on some 3 points  that located in the circle itself ^^

Example :

 A Circle x² + y² + Ax + By + C = 0 passed on 3 points (2,0) , (0, -2) , (4, -2) The Circle Equations is?

 

Insert the points to the x² + y² + Ax + By + C = 0 as (x,y)

(2,0) ---> 2² + 0² + A(2) + B.0 + C = 0

                 4 + 2A + 0 + C = 0 ........................ (1)

(0, -2) --> 0² + (-2)² + A.0 + B(-2) + C = 0

                             4 – 0 - 2B + C = 0 .......................... (2)

(4, -2) ---> 4² + (-2)² + A(4) + B(-2) + C = 0

                  16 + 4 + 4A - 2B +  C = 0

                  20 + 4A - 2B +  C = 0 ....................(3)

 
there are 3 quations and let's solve the equations!

Eliminate (1) dan (2)

4 + 2A + 0 + C = 0 .... (1) 

4 – 0 - 2B + C = 0 ..... (2)        

--------------------------------   -

0 + 2A + 2B + 0 = 0 .... (4)  

 

Eliminate (3) dan (2)

20 + 4A - 2B +  C = 0 ...(3)

4 – 0 - 2B + C = 0 ..... (2)    

------------------------------    -

 16 + 4A = 0

 

A = -4 -------> Substitute A as -4 to .... (4)

2A + 2B  = 0 ---> -8 + 2B = 0 --> B =4 

Substitute A and B as its value to equations number (1) / (2) / (3) 

4 + 2A + 0 + C = 0 .... (1) 
 4 – 8 + 0 + C = 0 

C = 4

after we found all the value A, B, C insert them to : x² + y² + Ax + By + C = 0 

So the Circle Equations  x² + y² - 4x + 4y + 4 = 0 ^^