Math Circles and spheres Circles in the plane

# Circles in the plane (2D)

With a centre and a radius, an equation of a circle can be set up in the plane.

The circle's cartesian equation is:

$(x-x_M)^2+(y-y_M)^2=r^2$
• $r$ is the radius
• $M(x_M|y_M)$ is the centre
!

### Remember

A circle in the plane is the set of all points $P$ that have the same distance $r$ (radius) from the centre $M$.

### Derivation

The length of the vector from $M$ to $P$ corresponds to the radius.

$|\vec{MP}|=r$

We can also represent this vector using the vectors $\vec{x}$ and $\vec{x_M}$ (see picture above).

$|\vec{x}-\vec{x_M}|=r$

If we square the whole equation, the absolute value is eliminated (calculation rule of vectors). We get the vector equation for circles in the plane.

$(\vec{x}-\vec{x_M})^2=r^2$

$\left(\begin{pmatrix} x \\ y \end{pmatrix}-\begin{pmatrix} x_M \\ y_M \end{pmatrix}\right)^2=r^2$

The upper coordinate equation is obtained by multiplying (scalar product).

The radius and centre can be read from the circular equation mentioned.

There are also other quadratic equations that describe a circle.

!

### Remember

Radius and centre can only be determined with the vector / cartesian equation. To change, use the completing the square.

### Example

Find the centre and radius of the circle $k$.

$k: x^2-4x+y^2+2y=20$

1. #### Completing the square

$x^2-4x+y^2+2y=20$

$x^2-4x\color{red}{+2^2-2^2}$ $+y^2+2y\color{blue}{+1^2-1^2}=20$

$(x-2)^2-2^2$ $+(y+1)^2-1^2=20$

$(x-2)^2+(y+1)^2-5=20\,\,|+5$

$(x-2)^2+(y+1)^2=25$

$k: (x-2)^2+(y+1)^2=25$
$k: (x-2)^2+(y+1)^2=5^2$
$M(2|-1)$ and $r=5$