How to shrink a circle to a point that is not its center

$$c(s,t)=(1-s)(x_0,y_0)+s(R \cos t,R \sin t)$$ 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.

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.