Parameterize a non-centered circle on origin in complex plane
How do I parameterize a circle centered at $2i$ on a complex plane? I know that we can parameterize a unit circle centered at origin as $C(t) = \cos(t) + i\sin(t)$ or $C(t)= \mathrm{Re}(it)$ but I'm not sure how to do it if not centered at the origin.
$\endgroup$2 Answers
$\begingroup$A circle of radius $R$ centered at the point
$c \in \Bbb C \tag 1$
is the set of all points
$z \in \Bbb C \tag 2$
which are a distance $R$ from $c$, that is the set of $z$ satisfying
$\vert z - c \vert = R; \tag 3$
this implies that $z - c$ is a complex number of modulus $R$, all of which may be expressed in the form $Re^{it}$, $t \in [0, 2\pi)$; thus (3) implies
$z - c = Re^{it}, \; t \in [0, 2\pi), \tag 4$
or
$z = c + Re^{it}, \; t \in [0, 2\pi); \tag 5$
taking
$c = 2i \tag 6$
as in the text of the question itself we find
$z = 2i + Re^{it}, \; t \in [0, 2\pi);, \tag 7$
giving a parametrization of the circle (3) with $c = 2i$ by $t \in [0, 2\pi)$.
We may express this in terms of real and imaginary parts of $z$ by writing
$z = 2i + Re^{it} = 2i + R\cos t + iR\sin t = R\cos t + (2 + R\sin t)i, \; t \in [0, 2\pi). \tag 8$
$\endgroup$ $\begingroup$A polar parametrization of a non-centered unit circle is shown in the following diagram
This shows that$$ r=a\cos(\theta)+\sqrt{1-a^2\sin^2(\theta)} $$parametrizes a unit circle from a point offset by $a$ from its center. Below is an animation showing this parametrization for $a\in[-1,1]$.
Comment on $\boldsymbol{a^2\gt1}$
Leaving out the angles where $a^2\sin^2(\theta)\gt1$, this works for circles that do not contain the origin. When $a\cos(\theta)\lt0$, the radius will be negative and so the points will be in the direction opposite from expected.
And Now For Something Completely Different...
While I was deriving the polar parametrization above, I made the following happy accident: the polar equation$$ r=\sqrt{1+a^2+2a\cos(2\theta)} $$for $a\in[-1,1]$:
