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

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 $$A$$.

• I think you meant "..conditions on the center $\;(h,k)\;$ of another circle..." – DonAntonio May 13 at 19:56
• @DonAntonio likely conditions are on radius of $B$, given the location of the new center – gt6989b May 13 at 19:57

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

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 $$1-d$$...

• No-sorry! I wanted conditions on $x$ and $y$ given a fixed radius of $B$--maybe I'll rephrase the question further. – Paul B. Slater May 13 at 20:13