A circle has a radius. On a cartesian plane, this radius becomes the hypotenuse of a right angle triangle, with one point lying at the centre of the circle (let's call this \(A(h, k\))), one lying on the circumference of the circle (let's call this \(B(x, y\))), and one situated at \(C(x, k)\).

The equation of a circle is essentially the pythagorean theorem applied to this triangle. Using the points we defined above, we would get something like \(radius^2=AC^2+BC^2\). This isn't particularly useful. We need the magnitude of \(AC\) and \(BC\). This is just the difference between \(x\) and \(h\) for the \(x\) magnitude; and the difference between \(y\) and \(k\) for the \(y\) magnitude.

We end up with \(r^2=(x-h)^2+(y-k)^2\).

A few things to note about this:

- If you set whatever is in either bracket equal to zero, you will get the coordinates of the center of the circle. If the coefficient of \(x\), for example, is 1, then you can simply invert the signs. I think you get the point.
- If you \(\sqrt{r}\), then you will get the magnitude of the radius.

Let's use \((x-3)^2+(y+4)^2=25\). Using the information above, we know that the center of this circle is located at \((3, -4)\) and its radius is 5.

Alternatively, let's use \((2x-4)^2+(y+8)^2=16\). Here, we know that the center of the circle is located at \(2, -8\), and its radius is 4.

However, we don't always deal with this format. Sometimes we end up with an equation like \(x^2+y^2=1-10x+2y\). Really dirty. When bringing this into the proper \((x-h)^2+(y=k)^2=r^2\) format, we always group together any like terms on one side and group any other constants on the other.

\(x^2+y^2=1-10x+2y\) becomes \(x^2+10x+y^2-2y=1\).

Now, we complete the square for the \(x\) and \(y\) terms individually.

\((x+5)^2-25+(y-1)^2-1=1\)

\((x+5)^2+(y-1)^2=27\)

This article was written on 02/10/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!