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enter image description here

The numbers are to identify the circles

I've came out with this list of 4 inequalities(1 each circle), but I don't know if this is the best method to calculate it, neither how to solve it:

$(x+\frac{d}{2})^2+y^2\geq r_1^2 \\ (x-\frac{d}{2})^2+y^2\geq r_2^2 \\ (x+\frac{d}{2})^2+y^2\leq r_{1'}^2 \\ (x-\frac{d}{2})^2+y^2\leq r_{2'}^2$

The radius of the big circles can't be smaller than the small's.

The center of $1'$ is equal to the center of $1$, and the same with 2s.

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  • $\begingroup$ I do not think that this problem has to be handled using inequalities. $\endgroup$ – Claude Leibovici Nov 27 '13 at 11:03
  • $\begingroup$ @ClaudeLeibovici The thing is that i want to calculate the area, yes. But i also want to take a formula for it, to graph it. That's why I choose inequalities. So, what is the good way to do it, then? $\endgroup$ – Arcotick Nov 27 '13 at 11:07
  • $\begingroup$ If all else fails, this looks doable (if laborious) with Green's theorem (in a form such as: the area of a region is the integral of $x\,dy$ over the boundary, traced counterclockwise), since each circle is easily parametrized, and the relevant angles for each boundary arc can be found in terms of $d$ and the radii. $\endgroup$ – Andrew D. Hwang Nov 27 '13 at 11:41
  • $\begingroup$ Let $B_1(\rho)$ and $B_2(\rho)$ be the balls with radius $\rho$ centered at $(-\frac{d}{2},0)$ and $(\frac{d}{2},0)$ respectively. Let $A(\rho_1,\rho_2)$ be the area of $B_1(\rho_1) \cap B_2(\rho_2)$. The area of the your shaped area is simply $$\frac12 \left( A(r_{1'}, r_{2'} ) - A(r_{1},r_{2'}) - A(r_{1'},r_{2}) + A(r_1,r_2) \right)$$ $\endgroup$ – achille hui Nov 27 '13 at 12:09
  • $\begingroup$ What are the known parameters? Do we know the radii of the circles? Otherwise this problem is impossible to solve; just knowing $d$ is insufficient. $\endgroup$ – YiFan Aug 19 '18 at 5:39
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WLOG, assume all centers lie on the $x$-axis

Firstly, we need to find the area bounded by $AIHN$, which can be determined by looking for the following areas:

$$A_{AIHN}=A_{\text{sector } ANI}+A_{\text{sector }ABI}-A_{\triangle ABI}\tag{1}$$ This requires you to solve for $\alpha$ and $\beta$, which I believe is difficult to derive without trigonometry.


Now we need to find the area bounded by $LPH$, which can be determined by:

$$A_{HLP}=A_{\text{sector }BHL}-A_{\triangle BHP}\tag{2}$$ Additionally, this requires you to solve for $\angle HBL=\gamma$.


From $(1)$ and $(2)$, you get half the shaded area (bounded by $HIK$):

$$A_{HIK}=A_{\text{sector }ABK}-A_{\triangle BKP}-(A_{AIHN}-A_{HLP})$$ Also, you again need to use trig to solve for $\angle ABK=\delta$

enter image description here


Verifying the answer

If you want to be sure, these are the following areas in algebraic terms.

Assume the following circles: $$C_{1',2'}\implies \left(x\pm k\right)^2+\left(y-h\right)^2=R^2\\ C_{1,2}\implies \left(x\pm k\right)^2+\left(y-h\right)^2=r^2$$

After some tedious work, you get the following areas from $(1)$:

$$A_{\text{sector }ABI}=\frac{1}{2} R^2 \cos ^{-1}\left(1-\frac{r^2}{2 R^2}\right)\\ A_{\triangle ABI}=\frac{1}{4} r \sqrt{4 R^2-r^2}\\ A_{\text{sector }ANI}=\frac{1}{2} r^2 \sec ^{-1}\left(\frac{4 k r}{4 k^2+r^2-R^2}\right)\\ A_{AIHN}=\frac{1}{2} r^2 \sec ^{-1}\left(\frac{4 k r}{4 k^2+r^2-R^2}\right)-\frac{1}{4} r \sqrt{4 R^2-r^2}+\frac{1}{2} R^2 \cos ^{-1}\left(1-\frac{r^2}{2 R^2}\right)$$

From $(2)$:

$$A_{\text{sector }LBH}=\frac{1}{2} r^2 \cos ^{-1}\left(\frac{-k^2+2 k r+r^2-1}{2 r^2}\right)\\ A_{\triangle HPB}=\frac{1}{2} k r \sqrt{1-\frac{\left(k^2-2 k r-r^2+1\right)^2}{4 r^4}}\\ A_{LPH}=\frac{1}{2} r^2 \cos ^{-1}\left(\frac{-k^2+2 k r+r^2-1}{2 r^2}\right)-\frac{1}{2} k r \sqrt{1-\frac{\left(k^2-2 k r-r^2+1\right)^2}{4 r^4}}$$

And finally, the shaded region:

$$A_{\triangle AKB}=\frac{\pi R^2}{6}\\ A_{\triangle PKB}=\frac{1}{4} \sqrt{3} k R\\ A_{HIK}=-\frac{1}{2} r^2 \sec ^{-1}\left(\frac{4 k r}{4 k^2+r^2-R^2}\right)+\frac{1}{2} r^2 \cos ^{-1}\left(\frac{-k^2+2 k r+r^2-1}{2 r^2}\right)-\frac{1}{2} k r \sqrt{1-\frac{\left(k^2-2 k r-r^2+1\right)^2}{4 r^4}}-\frac{1}{4} \sqrt{3} k R+\frac{1}{4} r \sqrt{4 R^2-r^2}-\frac{1}{2} R^2 \cos ^{-1}\left(1-\frac{r^2}{2 R^2}\right)+\frac{\pi R^2}{6}\tag{3}$$


Using calculus, the area of the shaded region should be: $$A_{HIK}=\int_0^{\frac{R^2-r^2}{4 k}} \left(\sqrt{-k^2-2 k x+R^2-x^2} -\sqrt{-k^2+2 k x+r^2-x^2}\right)\, dx\tag{4}$$


Now for $k=2, R=4,r=\sqrt5,h=0$, from $(3)$, we get:

$$\begin{align} A_{HIK}&=-2 \sqrt{3}+\frac{\sqrt{295}}{4}+\frac{8 \pi }{3}-1-8 \cos ^{-1}\left(\frac{27}{32}\right)+\frac{5}{2} \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)-\frac{5}{2} \sec ^{-1}\left(\frac{8}{\sqrt{5}}\right)\\ &=1.6151887775869405 \end{align}$$

From $(4)$ we get:

$$\begin{align} A_{HIK}&=\int_0^{\frac{11}{8}} \left(\sqrt{(2-x) (x+6)}-\sqrt{-x^2+4 x+1}\right) \, dx\\ &=\frac{8 \pi }{3}+\frac{1}{4} \left(-8 \sqrt{3}+\sqrt{295}-4-54 \sin ^{-1}\left(\frac{\sqrt{5}}{8}\right)-10 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)\\ &=1.6151887775869405 \end{align}$$

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  • $\begingroup$ your diagram is incorrect, the smaller circles SHARE SAME centers as larger ones $\endgroup$ – Randin Aug 19 '18 at 5:52
  • $\begingroup$ @RandinMichaelDivelbiss I overhauled my answer after overlooking the important detail $\endgroup$ – John Glenn Aug 19 '18 at 6:29
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Disclamer: This is a brute force approach and should not be used unless all else fails. A geometric solution is probably possible.

As per the suggestion by Andrew in the comments, we can use Green's Theorem to find the area. Green's Theorem states that for any curve $C$ in the $xy$-plane which bounds region $R$ and vector field $\vec{F}=\langle M,N\rangle$ which is differentiable in $R$, $$\oint_C \vec{F}\cdot d\vec{r} = \iint_R \left( \frac{\partial M}{\partial y}+\frac{\partial N}{\partial x} \right)dA. $$ To find the area of the region in the question, we just need a vector field where the integrand in the RHS is $1$. One such field is $\vec{F} = \langle 0,x\rangle$. We want to break $C$ (the boundary of the region whose area we are interested in) into four parts belonging to the four circles. Let $2k$ be the distance between the centers of the circles at the left and right side. The equations of these circles are $$ (x\pm k)^2+y^2=R^2 \quad \text{and}\quad (x\pm k)^2+y^2=r^2. $$ We need to parametrise the curves, and we will do so in terms of $y$. Solving for $x$ in each curve we get $$x=\pm k + \sqrt{R^2-y^2} \quad \text{and} \quad x=\pm k +\sqrt{r^2-y^2}$$ respectively. Let us cover $C$ counterclockwise starting from the bottom-most point, the intersection of the two big circles. Let us write $y_0$ for the $y$-coordinate of this point, $y_1$ for that of the intersection of a big and small circle, and $y_2$ for that of the intersection of the two small circles. Then the LHS can be written as: $$\begin{split} \oint_C Mdx + Ndy & = \oint_C xdy\\ & = \int_{y_0}^{y_1}k+\sqrt{R^2-y^2}dy + \int_{y_1}^{y_2} k+\sqrt{r^2-y^2}dy + \int_{y_2}^{y_1}-k+\sqrt{r^2-y^2}dy + \int_{y_1}^{y_0}-k+\sqrt{R^2-y^2}dy\\ & =^{\hspace{-0.3cm}*} 2k(y_2-y_0) +\int_{y_0}^{y_1}\sqrt{R^2-y^2}dy + \int_{y_1}^{y_2} \sqrt{r^2-y^2}dy - \int_{y_1}^{y_2}\sqrt{r^2-y^2}dy - \int_{y_0}^{y_1}\sqrt{R^2-y^2}dy\\ & = 2k(y_2-y_0). \end{split}$$ (Note that the expression after $=^{\hspace{-0.3cm}*}$ is obtained by isolating the $k$ from each integral and combining them after integrating.) It remains to find $y_0$ and $y_2$, which is quite easy. The intersections of the two big and small circles respectively clearly occur when $x=0$, so $$ y_0^2=R^2-k^2\implies y_0=-\sqrt{R^2-k^2} $$ and $$ y_2^2=r^2-k^2\implies y_2=-\sqrt{r^2-k^2}. $$ Combining these, we get that the area of the desired region is $2k(y_2-y_0)=2k\left(\sqrt{R^2-k^2}-\sqrt{r^2-k^2}\right)$. Since $2k=d$, this is the same as $$\mathrm{Area} = d\left( \sqrt{R^2-\frac{1}{4}d^2}+\sqrt{r^2-\frac{1}{4}d^2} \right).$$

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  • $\begingroup$ Thanks that took a good chunk out of your saturday night huh? ;) it will hopefully be the same as what I end up with in terms of unavoidable trig as in glenn solution .i wont post it till tommarow Zzzzzzz thanx again! $\endgroup$ – Randin Aug 19 '18 at 7:13
  • $\begingroup$ @RandinMichaelDivelbiss More like Sunday morning where I'm from. For more about Green's Theorem, you can see MIT's Multivariable Calculus lecture. $\endgroup$ – Lucas Aug 19 '18 at 7:23
  • $\begingroup$ Wow u go to mit Lucas ? $\endgroup$ – Randin Aug 20 '18 at 5:35
  • $\begingroup$ Of course not, but I find their open source lectures great reference material. $\endgroup$ – Lucas Aug 20 '18 at 7:12
  • $\begingroup$ Oh ok either way was a great answer $\endgroup$ – Randin Aug 20 '18 at 7:22

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