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
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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).


Determine radius and centre

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

There are also other quadratic equations that describe a circle.

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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

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

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

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

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

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

  2. Read off the centre and radius

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

    $k: (x-2)^2+(y+1)^2=5^2$

    Attention: The square root must first be removed for the radius! The center point is in the equation with the opposite sign!

    $M(2|-1)$ and $r=5$