Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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

1 Answer 1

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 $$\sqrt{r_1^2-1}+\sqrt{r_2^2-1} \leq \sqrt{(a_1-a_2)^2+(b_1-b_2)^2}$$

Proof:

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 $\sqrt{(a_1-a_2)^2+(b_1-b_2)^2} \geq r_1+r_2$ $\square$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.