1
$\begingroup$

I have a problem calculating the area of a circle segment. I know how to separate this into smaller tasks (triangle and remaining circle segment) that are basically easily solvable, but one distance is missing and I have no clue how to determine it. As you can see I need distance EB (which equals approx. 0.49597*r (by measuring)) to calculate the red area in the picture below:

Red: Area of interest

From there the triangle would be easy and the remaining segment can be calculated by using a formula I found on wikipedia:

formuala circle segment

This might be a really simple problem but all the people I talked to could not come up with a solution so I am really hoping somebody here can figure this out.

$\endgroup$
2
  • $\begingroup$ Please edit your question to include information about which points are known to you (given) and which you want to calculate... $\endgroup$
    – 5xum
    Jul 28, 2014 at 10:43
  • $\begingroup$ As you can see in the image above all points are solely depending on the radius r. The two circles have the same radius r and their centres are 2*r from one another. As you can also read in my question the only thing missing right now is the distance EB as all other distances can then be calculated from there. $\endgroup$
    – Daniel M
    Jul 28, 2014 at 10:48

2 Answers 2

2
$\begingroup$

You didn't give the radius. I assumed $2$.

The tangency point is at $(\sqrt2,\sqrt2)$, so that the center of the second circle is at $(2\sqrt2,2\sqrt2)$.

Its equation is $(x-2\sqrt2)^2+(y-2\sqrt2)^2=4$.

Intersecting with $y=2$ gives $x=2\sqrt2-\sqrt{4-(2-2\sqrt2)^2}\approx1.0080676825$ and $EB\approx0.9919323175$.

It will be easier to find the area by direct integration between $x_0$ and $x_1$ of the difference between the circle and the horizontal at $y_0=-\sqrt{4-x_0^2}$ (circle centered at the origin). $$A=\int_{x_0}^{x_1}(y_0+\sqrt{4-x^2})dx=(xy_0+\frac12\sqrt{4-x^2}+2\arcsin\frac x2)\Big|_{x_0}^{x_1}$$

$\endgroup$
1
  • $\begingroup$ thank you for your solution and the hint with the direct integration - unfortunately I can upvote your answer (reputation too low) $\endgroup$
    – Daniel M
    Jul 28, 2014 at 11:21
1
$\begingroup$

The way I read your construction, you have a square, a circle inscribed into that square and another circle of equal radius touching the first one and with its midpoint on one of the diagonals of the square. Correct me if this is wrong.

Let's choose $1$ as the radius for the circles. Then your second circle is centered at $(\sqrt2,\sqrt2)$ so it has the equation $(x-\sqrt2)^2+(y-\sqrt2)^2=1$. The point $E$ lies on that circle but also on the line $y=1$ so you plug these in and obtain

$$ (x-\sqrt2)^2+(1-\sqrt2)^2=1\\ x_{1,2}=\sqrt2\pm\sqrt{2\sqrt2-2} $$

Of these two solutions we take the left one, i.e. the smaller one, so the distance you want is

\begin{align*} \frac{\lvert EB\rvert}r&=1-x_1=1-\sqrt2+\sqrt{2\sqrt2-2}\\ \frac{\lvert EB\rvert}r&\approx 0.49596615875135963380702679143523906067070888158935 \end{align*}

This is not the approximation you state in your question, but it very much fits the picture: there the length you indicated is considerably shorter than the radii of the circles. Likely you got a factor of two error, because two times the number above is

$$2\frac{\lvert EB\rvert}r \approx0.99193231750271926761405358287047812134141776317870$$

$\endgroup$
1
  • $\begingroup$ Wow, it can be so simple :) Thank you very much. You are also right with the approximation as I forgot to add (0.99194)*r/2. $\endgroup$
    – Daniel M
    Jul 28, 2014 at 10:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .