How to shrink a circle to a point that is not its center Let's say we have a circle $C$ with a fixed radius $R$ on $\mathbb{R}^2$. An obvious parametrization to the circle would be:
$$C(t)=(R\cos(t),R\sin(t)) \\ t\in[0,2\pi]$$
I'm interested in shrinking the circle to a point. If we look at the limit where $R\to0$, it would be easy to see that the circle would shrink to the point $(0,0)$.
However, I'm trying to shrink the circle to another point. Let's say, for example, that $R=2$, and I want the circle to shrink to the point $P=(1,1)$. In order to do that, I have to make sure that through the process, the point $P$ would always be inside the area; For that reason, I must have another parametrization of the circle (one that would also help me to take a proper limit). 
Problem is I have no idea how to do that. I tried to draw a segment from the point $P$ to some generic point on the circle (and named it $r$), and then tried to parametrize the circle using $r$, but it was an algebraic challenge.
So I'd be very glad if you could help me with that. Also, if you thought of an easier solution - I would be glad to hear it too!
Thanks!
In short: Given the point $(x_0,y_0)$, find a parametrization $C(R,r,t)$ of a circle centered at $(0,0)$ with radius $R\geq\sqrt{x_0^2+y_0^2}$, such that $t\in[a,b]$ for some $a,b\in\mathbb{R}$, $r$ is fixed and is dependent on $R$ (which is constant), and:
$$\lim_{r\to0} C(R,r,t)=(x_0,y_0)$$
When $(x_0,y_0)$ is a point inside the circle, throughout the shrinking process.
 A: $$c(s,t)=(1-s)(x_0,y_0)+s(R \cos t,R \sin t)$$
Works, when $s=1$ one gets the circle and when $s=0$ one gets the point $P(x_0,y_0)$. For each $s$ the mapping $t\mapsto c(s,t)$ represents a circle of center $(1-s)(x_0,y_0)$ and radius $sR$. And $P$ is inside this circle if $P$ is inside the original circle.
A: Given the original circle
$\vec R(t) = (R\cos t, R\sin t), \tag 1$
and a point
$\vec P = (P_x, P_y) \tag 2$
to which we want to "shrink" the circle $\vec R(t)$, we may introduce the "shrinking parameter" $s$ and write
$\vec \phi(t, s) = (1 - s)\vec R(t) + s \vec P = \vec R(t) + s(\vec P - \vec R(t)), \; 0 \le s \le 1; \tag 3$
for $s = 0$, we have
$\vec \phi(t, 0) = \vec R(t), \tag 4$
the original circle, whilst when $s = 1$ we find
$\vec \phi(t, 1) = \vec R(t) + \vec P - \vec R(t) = \vec P; \tag 5$
as $0 \to s \to 1$, the point $\vec \phi(t, s)$ moves from the point $\vec R(t)$ on the given circle to the fixed point $\vec P$ along the line segment joining $\vec P$ with $\vec R(t)$; as such, $\vec \phi(t, s)$ for $s > 1$ clearly lies in the interior of the original circle $\vec R(t)$; this may be seen geometrically via the fact that the segment $\overline{ \vec P \vec R(t)}$ is contained in a chord of the circle $\vec R(t), 0 \le t < 2\pi$ when $\vec P$ lies inside the given circle.
