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Cartesian Equation of a Circle

Since a circle is made up of all the points a fixed distance (its radius) from a given point (its centre) then the equation of a circle simply needs to ensure this is true. This can be done using Pythagoras’s theorem. This is because we can draw a right angled triangle with the centre of the circle at one corner and the point on the circle at the opposite corner as shown below. The radius is then the hypotenuse, the vertical side is the difference between the y co-ordinate of the point and that of the centre and the horizontal side is the difference between the x co-ordinate of the point and centre. From Pythagoras we therefore know that a circle of radius r and centre (a,b) must have a Cartesian equation

r2 = (x-a)2 + (y-b)2

Circle on Cartesian axis

Circle on Cartesian axis

However, we can expand these brackets out to get

r2 = x2 – 2ax + a2 + y2 – 2by + b2

but since a2+b2+r2, -a and -b are all constant we can let

c = a2+b2-r2,
g = -a
f = -b

to get

x2 + y2 + 2gx + 2fy + c = 0
where the circle has a centre (-g,-f) and radius √(a2+b2-c2)

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  1. yyc
    July 11, 2009 at 4:37 am

    Slight error,
    x2 + y2 + 2gx + 2fy + c = 0 should be
    x2 + y2 + 2fx + 2gy + c = 0 otherwise it would expand into
    r2 = x2 – 2bx + a2 + y2 – 2ay + b2 instead of
    r2 = x2 – 2ax + a2 + y2 – 2by + b2

    Otherwise, thanks!

    • July 16, 2009 at 4:06 pm

      well done for noticing the mistake and thanks for telling me,
      i changed
      f=-a and g=-b
      to
      g=-a and f=-b
      instead to keep the final result the same
      thanks again for the comment and pointing out the mistake,
      david

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