distance between centres of two overlapping congruent circles

Question

If there are two overlapping congruent circles such that the area of intersection is 10% of the area of each circle, what is the distance between their centres in terms of the radius r cm?

Answer 1

Let $O$ and $O'$ be the centers and $P$ and $Q$ be the intersection points of both circles.

First, let's calculate the area of the circular segment $PQ$. We need the angle $\alpha=\angle POQ$.

Call $d=OO'/2$ and $r$ the radius of the circles. Then $$\alpha=2\arccos\frac dr$$

The area of the circular segment is obtained substracting the triangle $OPQ$ to the circular sector $OPQ$, and is $$\frac{\alpha r^2}{2}-d\sqrt{r^2-d^2}$$ being $\alpha$ expressed in radians (this makes the formulae clearer).

The area of intersection is twice the circular segment's. Therefore, you must solve this for $d$:

$$r^2\arccos\frac dr-d\sqrt{r^2-d^2}=0.05\pi r^2$$

I'd say that this is impossible to solve by algebraic methods, but any math program can do it with numerical methods, given the radius.

The equation becomes a bit cleaner if you call $k=d/r$ the ratio between $d$ and $r$. Dividing the equation by $r^2$ yields: $$\arccos k-k\sqrt{1-k^2}=0.05\pi$$ Sadly it still can't be solved by algebraic methods.

PS: Don't forget that the arccos must be in radians, and that $d$ is the distance between centers halved.

Answer 2

Is it pi(rrr/5-rrrr/200)? You assume one of the circles is xx+yy=rr while the other a circle with same radius but symmetric wrt x-axis. Assume one of the intersection points is (b,sqrt(rr-bb). Want b and then 2b is the required answer. The difference between the crcular sector between the x-axis and the point (b,sqrt(rr-bb) and the triangle (0,0),(b,0),(b,sqrt(rr-bb)) is 1/40 .pi.rr.

Answer 3

An uglier solution than the first answerer's would be:

$d=2r-\ell,$ where $r$ is the radius and $\ell$ is the overlap, given by its relation with $r$ from the equation below:

$$0=(r-\ell/2)\sqrt{\ell^2/4-r\ell}+r^2 \arcsin \left(\frac{r-\ell/2}{r}\right)-\frac{19}{20}\pi r^2$$

Using numerical method software (which I don't have) we can obtain $d$ in terms of $r$.

Answer 4

As ajotatxe answered, the solution of $$f(k)=\cos ^{-1}(k)-k\sqrt{1-k^2} -\frac{\pi }{20}$$ does not show analytical solution (remember that this is already the case for $x=\cos(x)$) and then numerical methods should be used.

The simplest is probably Newton, which, starting from a guess $k_0$, will update it according to $$k_{n+1}=k_n-\frac{f(k_n)}{f'(k_n)}$$ For this problem, $f'(k)=-2 \sqrt{1-k^2}$. You can also notice that $f(0)=\frac{9 \pi }{20}$ and that $f(1)=-\frac{ \pi }{20}$; a straight line going through these two points would be $\frac{9 \pi }{20}-\frac{ \pi }{20}k$ and it would intersect the $k$ axis for $k=0.9$ (this would be the first iterate of the secant method). So, let us start Newton method with $k_0=0.9$; so the successive iterates will be $$k_1=0.7871805391$$ $$k_2=0.8050335921$$ $$k_3=0.8053834963$$ $$k_4=0.8053836365$$ which is the solution for ten significant figures.

Being less lazy, we could have noticed that $\frac{\sqrt 3} 2 $ is close to the solution since $$f\big(\frac{\sqrt 3} 2\big)=\frac{7 \pi }{60}-\frac{\sqrt{3}}{4}\approx -0.0664936$$ A Taylor expansion built at this point would give $$f(k)=\left(\frac{7 \pi }{60}-\frac{\sqrt{3}}{4}\right)-\left(k-\frac{\sqrt{3}}{2}\right)+\sqrt{3} \left(k-\frac{\sqrt{3}}{2}\right)^2+O\left(\left(k-\frac{\sqrt{3}}{2}\right)^3\right)$$ Limiting to first order, this will give an approximate solution $$k_0=\frac{\sqrt{3}}{4}+\frac{7 \pi }{60}\approx 0.799532$$. Limited to second order, the approximate solution would be $$k_0=\frac{1}{30} \left(20 \sqrt{3}-\sqrt{5 \left(60-7 \sqrt{3} \pi \right)}\right)\approx 0.805812$$

More sophisticated (do not worry, you will learn soon about it) would consist in a Pade approximation of the function. Built at the same point, we should get for the simplest $$f(k)\simeq \frac{\frac{7}{60} \left(\sqrt{3} \pi -15\right) \left(k-\frac{\sqrt{3}}{2}\right)+\frac{1}{60} \left(7 \pi -15 \sqrt{3}\right)}{\sqrt{3} \left(k-\frac{\sqrt{3}}{2}\right)+1}$$ giving an approximate solution $$k_0=\frac{7 \pi -75 \sqrt{3}}{14 \left(\sqrt{3} \pi -15\right)}\approx 0.806399$$ Making the approximant slightly more complex, one could get $$k_0=\frac{6075 \sqrt{3}+1890 \pi -343 \sqrt{3} \pi ^2}{21600-840 \sqrt{3} \pi -392 \pi ^2}\approx 0.805173$$ As you can see (I hope), these approximations are not too bad.

Answer 5

Let $d$ be the distance between the centers of the circles.

Let $r$ be the radii of each circle.

Let $\theta$ be the angle subtended by the intersection of the two circles (the angle of the green triangle below at the center of the blue circle).

We can find the area of the intersection $A$ as twice the area of the black region ($2 \cdot A_B$) depicted above. The black region is known as a circle segment.

We can calculate $A_B$ by finding the area of the circle sector (the addition of the green and black regions) $A_S$ and subtracting the area of the green triangle, $A_G$.

---

I will express $A$ as a function of $\theta$ first, then after that I will substitute $\theta(d)$.

It is simple to see that $d(\theta)/2 = r \cos(\theta/2)$, and consequently, $\theta(d) = 2 \arccos \left( \frac{d}{2r} \right) $

Finding $A(\theta)$:

\begin{align} \frac{A}{\pi r^2} &= \frac{2A_B}{\pi r^2} \\ &= \frac{2}{\pi r ^2} (A_S - A_G) \\ &= \frac{2}{\pi r ^2} \left(\frac{\theta}{2\pi} \cdot \pi r^2 - A_G \right) \\ &= \left( \frac{\theta}{\pi} - \frac{2 A_G}{\pi r^2}\right) \\ &= \left( \frac{\theta}{\pi} - \frac{2}{\pi r^2} \left[ r^2 \cos(\theta/2) \sin(\theta / 2)\right]\right) \\ &= \frac{1}{\pi} \left( \theta - 2 \cos(\theta/2) \sin(\theta /2) \right) \end{align}

Since I've divided $A$ by $\pi r^2$, we are going to solve later for a value of $0.1 = 10\%$ of the total area of one of the circles.

Finding $A(d)$:

Now it is a simple matter of substitution.

\begin{align} \frac{A}{\pi r^2} &= \frac{2}{\pi} \left( \arccos \left( \frac{d}{2r} \right) - \underbrace{\cos \left(\arccos \left( \frac{d}{2r} \right) \right)}_{d/2r} \cdot \underbrace{\sin \left(\arccos \left( \frac{d}{2r} \right) \right)}_{\sqrt{1-d^2/4r^2}}\right) \\ &= \frac{2}{\pi} \left( \arccos \left( \frac{d}{2r} \right) - \frac{d}{2r}\sqrt{1-\frac{d^2}{4r^2}} \right) \end{align}

Final answer:

I do not believe that it is possible to rearrange this last equation to make $d$ the subject. So to solve, we must use numerical, non-exact methods. I have done so using Mathematica to find the answer in your case when $A/(\pi r^2) = 0.1$ of the area of one of the circles:

$$\frac d r \approx 1.61077$$