# How to derive the formula for a circle from scratch

If you’d like to derive the formula for a circle from absolute scratch, then your best option would be to draw a diagram such as the one below:

If you look at this diagram carefully, what you will notice is:

• A circle exists and each point on this circle has the coordinate (x, y).
• The centre of the circle can be found at (a, b).
• The circle has a radius ‘r’.
• The right angled triangles in the diagram each have an adjacent length, opposite length and hypotenuse (r).

Once you’ve prepared a similar diagram, your next aim should be to turn your attention towards the right angled triangles which exist within the circle. You should also think about the many different right angled triangles which could fit within the circle provided they emanate from the centre point (a, b).

The reason I’ve mentioned these right angled triangles is because according to Pythagoras’ theorem, when you have a right angled triangle – its adjacent length squared plus its opposite length squared is equal to the length of its hypotenuse squared:

Now, in this case – the adjacent lengths of the right angled triangles which can fit within the circle on the diagram can be described using the expression:

$\left( x-a \right)$ or $\left| x-a \right|$

The opposite lengths can be described using the expression:

$\left( y-b \right)$ or $\left| y-b \right|$

Also, very interestingly:

• Each of the right angled triangles you can think of has a hypotenuse ‘r’.
• ${ \left( x-a \right) }^{ 2 }={ \left| x-a \right| }^{ 2 }$
• ${ \left( y-b \right) }^{ 2 }={ \left| y-b \right| }^{ 2 }$

When you combine all the information above, what you get is a neat formula which looks like this:

${ \left( x-a \right) }^{ 2 }+{ \left( y-b \right) }^{ 2 }={ r }^{ 2 }$

And it turns out… This is the formula for a circle on the x, y plane, whereby, (a, b) is the centre of the circle and ‘r’ is the length of its radius. How spectacular is that? 🙂