Given a circle $A$ of area 1 centered at $\{0,0\}$, give conditions that another circle $B$ of known area <1, lies totally within $A$

Question

Given a circle $A$ of area 1 centered at $\{0,0\}$--so, of radius $\frac{1}{\sqrt{\pi}}$--give conditions on the possible location of the center $\{x,y\}$ of another circle $B$ of known area $\pi r^2 <1$, and thus of known radius $r < \frac{1}{\sqrt{\pi}}$--so that it lies totally within $A$.

Answer 1

If $B$ is a circle such that $x^2 + y^2 + 2gx + 2fy + c=0$ then it has a center $(-g,-f)$ and a radius $r = \sqrt{g^2 + f^2 - c} $

if radius of $B$ = $r$ is known then, for $B$ to be completely inside $A$;

(distance between centers of A,B) + (radius of $B$) $\le$ (radius of $A$) $$\sqrt{g^2 +f^2} + r \le \frac{1}{\sqrt{\pi}} $$ $$\sqrt{g^2 +f^2} \le \frac{1}{\sqrt{\pi}} - r $$

as $(-g,-f)$ are just coordinates of the center of $B$ generally as $(-g,-f)$ $\to$ $(x,y)$ we have; $$\sqrt{x^2 +y^2} \le \frac{1}{\sqrt{\pi}} - r $$

as $r \lt \frac{1}{\sqrt{\pi}}$ we have; $${x^2 +y^2} \le {(\frac{1}{\sqrt{\pi}} - r})^2 $$

Answer 2

I am assuming you want conditions on the radius of $B$, given that the center of $B$ is at the point $(x,y)$.

The only way to keep $B$ entirely inside $A$ is to restrict the radius of $B$ so that its boundary touches $A$ on the closest point.

That point will lie on the line through the center of $A$ and the center of $B$. So we find the offset of the center of $B$ to be $$d = \sqrt{x^2+y^2}$$ long, hence the radius of $B$ must be at most $1-d$...