Posts Tagged ‘circle’

Cartesian Equation of a Circle

June 28, 2009 2 comments

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)


Area And Circumference of a Circle : pi

March 2, 2008 58 comments

This site is now at

This is a basic guide to using pi to find the area and circumference of a circle using pi. And also explores why pi makes our formulae work.

Circle radius and circumference

area =πr2

circumference = 2πr or πd

where r = radius and d=diameter


First lets look at the area of a circle, given by area =πr2. This is simple enough to use, we multiply the radius by itself and then by pi.
Does this make sense?
Well r squared is at least going to be an area but it might be a bit small so we multipy by pi. However this doesn’t explain much until we consider what pi is, the easiest way i find to do this is as follows
If we imagine a square that the circle fits inside perfectly(so it touches all four sides like the one above) r squared would give us one quadrant, so the area of that square is 4 x r2 . Of course the circle’s area is a bit smaller so we need to find the ratio between the areas of the square and circle. If we then times this value by four we have a magic constant to multiply r squared by to find the area of a circle (we times by four because we need the area of 4 quadrants and r squared gives us one).
Now this magic constant is pi (which makes sense being just over 3, meaning the area of the circle is just over 3/4 of the area of the square).
The circumference of a circle is given by 2πr or πd. This seems simple, we just multiply the diameter (2r) by our magic constant pi.
Does this also make sense?
seeing as we only have one r this time so only one length it seems we are just finding a factor to increase the length by to make a different length(the circumference) which makes sense.
Again lets consider the square into which our circle fits perfectly, the perimeter of this square would be 4 time the length of one of the sides.
Now the length of the sides = the diameter so the perimeter is 4d.
Notice again that the value we are trying to find for the square is multiplied by 4, but for a circle were going to need a ratio that is a bit smaller.
So we need to replace the 4, for a square, with another, smaller, number — it seems pi will do the job.
To me when I consider pi I don’t look at it as a magical fundamental constant, but more a magical fundamental constant multiplied by four, because when I consider how these formula work using pi this is how they seem to work.
So this new constant is really the ratio of
area of square to area of circle
perimeter of square to circumference of circle.
and it = pi/4 = 0.785398….
so if you have a value for a square and you want a similar value for the circle you just need to multiply it by this number and you’ll have your answer 🙂
I welcome comments, improvements or errors in this post. Please leave your comments below or email me at

By David Woodford