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I have two disks $(x-a_1)^2+(y-b_1)^2\leq r_1^2$ and $(x-a_2)^2+(y-b_2)^2\leq r_2^2$, where $a_1$, $b_1$, $r_1$, $a_2$, $b_2$, $r_2$ are all known. What kind of constraint can I put on $a_i$, $b_i$ and $r_i$ that the intersection of two disks is inside unit circle? The question is for intersection of two disks, but the generalization for $n$ disks would be even better.

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unit circle centered at any particular point? – Sasha Sep 2 '12 at 3:51
I guess it doesn't matter, but, maybe you want $r_1^2$ and $r_2^2$ for the sake of geometric clarity. – James S. Cook Sep 2 '12 at 3:51
@Sasha, i have to define $a_i$, $b_i$ and $r_i$, so i think they can be centered any point – kotoll Sep 2 '12 at 3:54
@JamesS.Cook equations are edited. – kotoll Sep 2 '12 at 3:55
You might be able to simplify the problem with a well-chosen Möbius transformation. Maybe try transforming the unit circle into a line, or transform the two disks so that they have the same radius. – Chris Culter Jul 29 '13 at 6:36

Lemma: For positive $a$ and for $0 \leq r < 1$, let $\rho(a,r)$ be the radius of the smallest disk containing $D_1\cap D_2$, where $D_1 = \{x^2+y^2 \leq 1\}$ and $D_2 = \{(x-a)^2 + y^2 \leq r^2\}$. Then $\rho(a,r) \leq R \in [0, 1]$ iff either $r \leq R$, both $r > R$ and $\sqrt{1-R^2}+\sqrt{r^2-R^2} \leq a \leq 1+r$, or $a \geq 1+r$.

Proof: We can then identify three cases:

  • Case 1: If $a \leq \sqrt{1-r^2}$, then a whole diameter of $D_2$ is contained in $D_1 \cap D_2$, so $\rho(a,r) = r$. Therefore $\rho(a,r) \leq R$ holds when $r \leq R$.

  • Case 2: If $a \in (\sqrt{1-r^2},1+r)$, then $\rho(a,r)$ is half of the length of the line segment connecting the points of $\partial D_1 \cap\partial D_2$, which is $\sqrt{1-\left({a^2+1-r^2 \over 2a}\right)^2}$. $\sqrt{1-\left({a^2+1-r^2 \over 2a}\right)^2} \leq R$ simplifies to $a^2 -2\sqrt{1-R^2}a + 1-r^2 \geq 0$. This is a quadratic in $a$ with discriminant $\Delta = 4(r^2-R^2)$, so if $r \leq R$ then the inequality always holds. If $r > R$, then we must have $a \geq \sqrt{1-R^2} + \sqrt{r^2-R^2}$ (the other formal solution is discarded since $a > \sqrt{1-r^2}$).

  • Case 3: If $a \geq 1+r$ then the two disks are disjoint or intersect in a point, so $\rho(a,r) = 0$. Therefore $\rho(a,r) \leq R$ holds for all $R$. $\square$

Theorem: The intersection of the disks $(x-a_1)^2+(y-b_1)^2 \leq r_1^2$ and $(x-a_2)^2+(y-b_2)^2 \leq r_2^2$ is contained in a unit circle iff either (a) $r_1$ and/or $r_2$ are at most one, or (b) both $r_1$ and $r_2$ are greater than one and $$\sqrt{r_1^2-1}+\sqrt{r_2^2-1} \leq \sqrt{(a_1-a_2)^2+(b_1-b_2)^2}$$


Assume without loss of generality that $r_1 \geq r_2$. Using the transform $(x,y)\to(r_1x+a_1, r_1y+b_1)$ yields the disks $$x^2+y^2 \leq 1;\,\,\,\,\, (x+\hat{a})^2+(y+\hat{b})^2 \leq \hat{r}^2$$ where $\hat{a}={a_1-a_2 \over r_1}$, $\hat{b}={b_1-b_2 \over r_1}$, and $\hat{r}={r_2 \over r_1}$. The problem becomes finding the constraints that guarantee that the intersection of these disks in contained in a disk of radius $R=1/r_1$. Letting $\hat{c}=\sqrt{\hat{a}^2+\hat{b}^2}$, we see this is equivalent when looking at the intersection of the disks $$x^2+y^2 \leq 1;\,\,\,\,\, (x-\hat{c})^2+y^2 \leq \hat{r}^2$$ By the previous lemma, this means that either

  • $\hat{r} \leq R$: This is equivalent to $r_2 \leq 1$.

  • Both $\hat{r} > R$ and $\sqrt{1-R^2}+\sqrt{\hat{r}^2-R^2} \leq \hat{c} \leq 1+\hat{r}$: This is equivalent to $r_2 > 1$ and $\sqrt{r_1^2-1}+\sqrt{r_2^2-1} \leq \sqrt{(a_1-a_2)^2+(b_1-b_2)^2} \leq r_1+r_2$.

  • $\hat{c} \geq 1+\hat{r}$: This is equivalent to $\sqrt{(a_1-a_2)^2+(b_1-b_2)^2} \geq r_1+r_2$ $\square$

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