# The Equation of a Circle

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 $$r^2=(x-h)^2+(y-k)^2$$.

1. 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.
2. 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!