Ecuacion General De La Circunferencia

Mathematics III
“General equation of the circumference with center outisde the origin”.

Faculty: I.S.C.

Team members:
(.....)

October 17th, 2012

General equation of the circumferencewith center outisde the origin.
What is the circumference?
It`s a curve closed flat, where its points are intersect in the center

Where does the general equation of the circumference with centeroutside the origin go out?
d= √(X2-X1)2 + (Y2-Y1)2
d2=√(X2-X)2+ (Y2-Y1)2
d=r
r2= (X2- X1)2+ (Y2- Y1)2
c= (X1, Y1) P=(X2, Y2)
h, k=c
r2= (X2-h)2+(Y2-k)2
The general formula is:
(x-h)²+(y-k)²=r²

Where:
(x, y) is: any point in the circumference.
(h, k) / (a, b) are: the coordinates of center.
r is: circle radius.
Representation:

The result of the general formula in the circle withcenter outside the origin must take the following structure:
x2  + y2 + Dx + Ey + F = 0

For example:
Get the general equation of the circle with center (2, -3) and r = 5.
C (2, ─3),
r = 5 Method developed:
We know: C (2, ─3), r = 5
Ordinary formula will;
(x ─ a)2  +  (y ─ b)2 = r2 
(a, b): C(2,-3) (center coordinates)
Also represented with other variants, but you can use either;(x - h)2 + (y - k)2 = r2
(h, k): C(2,-3) (center coordinates)

Remain as ordinary equation:
(x ─ 2)2  +  (y ─ ─ 3)2  = 52
(x ─ 2)2  +  (y + 3)2  = 52
(x ─ 2)2  +  (y + 3)2  = 25
We have theordinary equation:
(x ─ 2)2  +  (y + 3)2  = 25
Develop its two binomials:
(x  ─ 2) (x  ─ 2) + (y  +  3) (y  +  3) = 25
(x2 ─ 2x ─ 2x + 4) + (y2 + 3y + 3y + 9) = 25
(x2 ─ 4x  +  4) + (y2 + 6y +9) = 25
We order our equation according to the general structure:
x2  +  y2  ─ 4x  +  6y + 4 + 9 ─ 25 = 0
x2  +  y2  ─ 4x  +  6y  ─ 12 = 0

It is the general equation of the circumference withcenter in the outside the origin coordinates C= (2, ─3) y and r=5.

2.- Find the center, radius and graph of a circumference given its ordinary equation:
(x-5)2+ (y-3)2=16
So the formula is:...