Ask Me Help Desk - How to find the distance between two overlapping circles?

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How to find the distance between two overlapping circles?

I have the following question in the ‘radians’ section of my coursework:
An eclipse of the sun is said to be 10% total when 10% of the area of the sun’s disc is hidden behind the disc of the moon.
Given the radii of both is r.
Calculate, in terms of r, the distance between the centres of the two discs.

I assume I need to obtain a value for theta.
I have gotten as far as
theta - sin theta = pi/20 but then I am stumped.

The answer is 1.61r

Any help would be appreciated. Thanks.

Quote:

Originally Posted by AdrianCavinder

I have the following question in the ‘radians’ section of my coursework:
An eclipse of the sun is said to be 10% total when 10% of the area of the sun’s disc is hidden behind the disc of the moon.
Given the radii of both is r.
Calculate, in terms of r, the distance between the centres of the two discs.

I assume I need to obtain a value for theta.
I have gotten as far as
theta - sin theta = pi/20 but then I am stumped.

The answer is 1.61r

Any help would be appreciated. Thanks.

I don't think it's possible to solve this in closed form. Using calculus (or looking up the formula for area of a circular segment), you find that the area is a transcendental equation in d (distance between the centers). In other words, you have to solve it numerically.

You can find the formula on this page. For this particular case, the equation works out to:

$(0.10)\pi r^2=2r^2 \cos^{-1}$ \frac{d}{2r} $ - \frac{d}{2}\sqrt{4r^2-d^2}$

Solving numerically for d does, indeed, give you $d=1.61r$ as an answer.

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I solved it different way, but arrived at the same conclusion that there is no closed-form solution. The difference between the area of the circle segment ABC and the triangle formed by ACD is 1/4 the area of overlap of the circles:

$ R^2 \frac { \theta} 2 - \frac 1 2 R^2 \sin \theta \cos \theta = \frac {\pi R^2} {40}\\ \frac \theta 2- \frac 1 2 \sin \theta \cos \theta = \frac \theta 2 - \frac 1 4 \sin 2 \theta = \frac {\pi} {40 $

Solve for theta using numerical techniques and you get theta = 0.63448 radians.

The distance between the centers of the circles is two times length AD:

$ x = 2 R \cos \theta = 1.61R $

Quote:

Originally Posted by ebaines

I solved it different way, but arrived at the same conclusion that there is no closed-form solution. The difference betweeen the area of the circle segment ABC and the triangle formed by ACD is 1/4 the area of overlap of the circles:

$ R^2 \frac { \theta} 2 - \frac 1 2 R^2 \sin \theta \cos \theta = \frac {\pi R^2} {40}\\ \frac \theta 2- \frac 1 2 \sin \theta \cos \theta = \frac \theta 2 - \frac 1 4 \sin 2 \theta = \frac {\pi} {40 $

Solve for theta using numerical techniques and you get theta = 0.63448 radians.

The distance between the centers of the circles is two times length AD:

$ x = 2 R \cos \theta = 1.61R $

Hi. Thanks. Very helpful. Based on the information you gave, I went back to my work and found the answer, but by using a slightly different approach.
During this section of the course we have been given two formulae for finding the area of a chord segment and for finding the length of the chord segment. So our answer can really only be based on these two formulae which are
Area of a circular sector = 1/2r^2theta and
Length of the arc of a circular sector = rtheta
Given that the area of the triangle created between the chord and centre of the circle is 1/2r^2sintheta, then the area of the chord is
1/2r^2theta - 1/2r^2sintheta
This would lead me to 1/2r^2theta - 1/2r^2sintheta = pir^2/10
which reduces to theta - sin theta = 0.31416(pi/10)
When I got there last time, that’s when I got stumped because how on earth do you evaluate theta - sin theta?
Now I assume (please correct me if I’m wrong) that what you mean by ‘numerical techniques’ is somewhat like trial and error?
If I apply a trial and error technique to the equation, I get theta = 1.26896, so theta/2 = 0.63448
Then the rest is the same.
Thanks so much. The only clarification I would appreciate is whether I’m right in my assumption about ‘numerical techniques’. Oh, and may I ask what you mean by ‘no closed-form solution’?

Quote:

Originally Posted by jcaron2

I don't think it's possible to solve this in closed form. Using calculus (or looking up the formula for area of a circular segment), you find that the area is a transcendental equation in d (distance between the centers). In other words, you have to solve it numerically.

You can find the formula on this page. For this particular case, the equation works out to:

$(0.10)\pi r^2=2r^2 \cos^{-1}$ \frac{d}{2r} $ - \frac{d}{2}\sqrt{4r^2-d^2}$

Solving numerically for d does, indeed, give you $d=1.61r$ as an answer.

Hi. Thanks. Very helpful. The information you gave was a bit too technical for the level that is being taught (this is a AS level mathematics course - 11th grade). But I went back to my work and found the answer, but by using a slightly different approach.
During this section of the course we have been given two formulae for finding the area of a chord segment and for finding the length of the chord segment. So our answer can really only be based on these two formulae which are
Area of a circular sector = 1/2r^2theta and
Length of the arc of a circular sector = rtheta
Given that the area of the triangle created between the chord and centre of the circle is 1/2r^2sintheta, then the area of the chord is
1/2r^2theta - 1/2r^2sintheta
This would lead me to 1/2r^2theta - 1/2r^2sintheta = pir^2/10
which reduces to theta - sin theta = 0.31416(pi/10)
When I got there last time, that’s when I got stumped because how on earth do you evaluate theta - sin theta?
Do I assume that by saying yo have to solve this ‘numerically’, that that means there is a certain amount of ‘trial and error’ involved?
If I apply a trial and error technique to the equation, I get theta = 1.26896, so theta/2 = 0.63448
Then 2rCostheta = the distance between the 2 centres which = 1.61r
Thanks so much. The only clarification I would appreciate is whether I’m right in my assumption solving ‘numerically’. Oh, and may I ask what you mean by ‘closed-form’?

Hello Adrian: yes, "numerical techniques" means using a technique that applies a series of guesses to zero in on the answer. You don't get an exact "closed form" solution (meaning you can't get to a formula for "theta = ...."), but by making guesses with ever smaller amounts of error you can get as close to the "true" answer as you want. There are several different techniques that can zero in on the answer pretty quickly, suitable even for running on an Excel spreasdheet. But for now simply think of it as (1) make a guess and see what the erreor is, then (2) make another guess and see what the new error is, then (3) make a third guess based on what you learned from the errors from guesses 1 and 2. Repeat until the error gets sufficiently small.

Quote:

Originally Posted by ebaines

Hello Adrian: yes, "numerical techniques" means using a technique that applies a series of guesses to zero in on the answer. You don't get an exact "closed form" solution (meaning you can't get to a formula for "theta = ...."), but by making guesses with ever smaller amounts of error you can get as close to the "true" answer as you want. There are several different techniques that can zero in on the answer pretty quickly, suitable even for running on an Excel spreasdheet. But for now simply think of it as (1) make a guess and see what the erreor is, then (2) make another guess and see what the new error is, then (3) make a third guess based on what you learned from the errors from guesses 1 and 2. Repeat until the error gets sufficiently small.

Okay, thanks, that helps to clarify. I really thought I was off thinking that I would need to use the ‘numerical technique’, although I would have thought that if theta - sin theta is defined, then there should be a formula to evaluate theta. Anyway, your help is very much appreciated. I actually had another question from the same course which I posted a while back, If you don’t mind, I’ll just copy it here and if you have an explanation, I’d be very thankful. I’m actually teaching this course here in India for the first time, so I need to know the ‘why’ to every question.
I am really stumped by this question. It’s in the AS level course (11th grade).
An infinite geometric series has first term a and sum to infinity b, where b ≠ 0. Prove that a lies between 0 and 2b.
How do I go about proving this? I can find the sum to infinity in terms of a, b and r, but not sure if I’m heading in the right direction. Any help would be appreciated.

$\sum_{n=0}^\infty ar^n = b$

Since the series converges, you know that r must lie between -1 and 1 (that's a precondition of convergence for a geometric series).

Given this condition, we know that the series converges to:

$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$

Thus, we can say

$\frac{a}{1-r}=b$

or

$a=(1-r) \cdot b$ ,

where $-1 \leq r \leq 1$

Plugging in the two extremes for r (-1 and 1), you'll see that the possible values for a are bounded by 0 and 2b, with intermediate values of r falling somewhere in between.

Quote:

Originally Posted by jcaron2

$\sum_{n=0}^\infty ar^n = b$

Since the series converges, you know that r must lie between -1 and 1 (that's a precondition of convergence for a geometric series).

Given this condition, we know that the series converges to:

$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$

Thus, we can say

$\frac{a}{1-r}=b$

or

$a=(1-r) \cdot b$ ,

where $-1 \leq r \leq 1$

Plugging in the two extremes for r (-1 and 1), you'll see that the possible values for a are bounded by 0 and 2b, with intermediate values of r falling somewhere in between.

Excellent. Thanks so much. Very clear explanation. On the same subject, I had another question that I had difficulty with. Any help would due appreciated.
I had to show that the sum of the infinite series 1 - x + x^3 - x^4... is equal to 1/(1 + x + x^2) and state the values of x for which this is valid. I found the first part with no difficulties, but I’m unsure how to find the values for x. The answer states that mod x is less than 1, but I have no idea how they came up with that answer. Any help would be appreciated.

[Disregard this until you read my EDIT at the bottom]

You really need to ask this question in a new thread. These forums are made for people to be able to search at a later point in time. They'll see your original question about distance between overlapping circles and never guess that there are questions and answers regarding infinite series further into the thread.

Meanwhile, I'll copy my answer to the previous question about the geometric series into the thread from a few weeks ago when you originally asked the question.

EDIT: Sorry, I just saw that you already did ask this in a separate thread. I'll answer there.

How to solve 2theta-sin2thetaequals pi divided by 10 by numerical method

You're trying to solve:

$ 2 \theta - \sin(2 \theta) = 10 $

Are you familiar with Newton's method? First rearrange equation so the right hand side equals zero, and your task is to find a value for theta such that $f(\theta)=0$ . Make an initial guess for the value of $\theta$ - call it $\theta _1$ - and plug it into $f(\theta)$ . (Hint - you should be able to see that the answer is somewhere around 5, since the $sin(2 \theta)$ varies between +/- 1, so $2\theta$ must be between 9 and 11.) The result will most likely not be 0, but rather some non-zero value which we will call the "error" or e_1. Now calculate the value of the derivative of $f(\theta_1)$ , call it $f'(\theta_1)$ . For your next guess you want to extrapolate the slope of $f(\theta)$ to the y axis and see what he resulting new value of $\theta$ is. To do this set:

$ \theta_2 = \theta_1 - \frac {e_1} {f'(\theta_1)} $

This gives you a new guess for theta. Continue this process calculating errors and slopes until the value of the error becomes "reasonably small."

Post back with what you find.

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