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# Area of intersecting circles

+13
2247
10
+28076

Here's a puzzle for when the site activity is low!

In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown.  What is the area of their intersection (i.e. the shaded region)?

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Oct 28, 2015
edited by Alan  Oct 28, 2015

### Best Answer

#6
+22586
+33

Here's a puzzle for when the site activity is low!

In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown.  What is the area of their intersection (i.e. the shaded region)?

$$circle_1: x^2 + y^2 = 4 \qquad r_1 = 2\\ circle_2: (x-2)^2 + (y-2)^2 = 1 \qquad r_2 = 1\\$$

intersections $$S_1 \text{ and } S_2$$:

$$2 x_{s}^2 - 2a \cdot x_s + a^2 - r_1^2 = 0 \qquad a = \frac{3r_1^2-r_2^2}{2r_1} = \frac{11}{4}\\ \vec{S_1} =\dbinom{ \frac18(11-\sqrt{7}) }{ \frac18(11+\sqrt{7}) }\\ \vec{S_2} =\dbinom{ \frac18(11+\sqrt{7}) }{ \frac18(11-\sqrt{7}) }$$

sectors angels (cosinus-rule) $$\alpha_1 \text{ and } \alpha_2$$:

$$\cos{(\alpha_1)} = \frac{57}{64} \qquad \alpha_1 = 27.0481105464^{\circ}\\ \cos{(\alpha_2)} = \frac{9}{16} \qquad \alpha_2 = 55.7711336722^{\circ}\\$$

Areas sectors $$A_{s_1} \text{ and } A_{s_2}$$:

$$A_{s_1} = 4\pi \frac{27.0481105464^{\circ}}{360^{\circ}}\\ A_{s_2} = \pi \frac{55.7711336722^{\circ}}{360^{\circ}}\\$$

Areas triangles $$A_{t_1} \text{ and } A_{t_2}$$:

$$A_{t_1} = \frac{1}{2} \cdot \left| \vec{S_2}\times \vec{S_1} \right| = \frac{11}{32}\sqrt{7}\\ A_{t_2} = \frac{1}{2} \cdot \left| \left[\vec{S_1}-\binom{2}{2}\right] \times \left[\vec{S_2}-\binom{2}{2}\right] \right|= \frac{5}{32}\sqrt{7}$$

$$\text{The area of their intersection }\quad A = A_{s_1}-A_{t_1}+A_{s_2}-A_{t_2}$$

$$A=\frac{\pi}{360}(4\cdot 27.0481105464^{\circ}+55.7711336722^{\circ})-\frac{ \sqrt{7} }{2}\\ A= 1.43085212603-1.32287565553\\ \mathbf{A=0.10797647050}$$

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Oct 28, 2015

### 9+0 Answers

#2
+101952
+7

I could have solved this with some messy Algebra, but....I decided to cheat....LOL!!!

Placing the center of the larger circle at the origin, we have these two equations :

x^2 + y^2 = 4      and  (x - 2)^2 + (y - 2)^2 = 1

Here's a diagram :

The area of triangle ABC  =  2*sin(27.048)  = about .909 units^2

And the area of the sector ABC = about 944 units^2

Similarly, the area of triangle  DAB = (1/2)sin(55.771)  = about  .413  units^2

And the area of sector DAB = about .4867 units^2

So...the approximate total area  of the  area of intersection = [ .944 - .909] +  [ .4867 - .413] = about .1087 units^2

OOPS....you are correct Alan....I have provided an edit......duh!!!....I think I mis-placed my decimal point.....Is my answer now correct???

Oct 28, 2015
edited by CPhill  Oct 28, 2015
#3
+28076
+5

Good try Melody, but this shows why it isn't correct:

Will now have a look at Chris's result (which still doesn't match what I'm expecting!)

Ah! If Chris could do adds and take aways correctly he would have the right answer!

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Oct 28, 2015
edited by Alan  Oct 28, 2015
#4
+102486
0

Yes I know that Alan - That is why I wrote that it was not correct.

I suppose i should jut have deleted it.  I will delete it now.

I am working on a solution that uses co-ordinate geometry.  I am sure I can do it that way but it is quite painful especially if I want to keep the answer exact.  ://

There you go Chris, Yours is wrong too.   LOL

But your diagram is fantastic.

I am only joking Chris,  You answer is great too :))

Thanks for this distraction Alan.  I have enjoyed it.

Oct 28, 2015
edited by Melody  Oct 28, 2015
edited by Melody  Oct 28, 2015
edited by Melody  Oct 28, 2015
#5
+28076
0

I find it fascinating that the question is much more complicated than it appears to be at first sight.

Oct 28, 2015
#6
+22586
+33
Best Answer

Here's a puzzle for when the site activity is low!

In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown.  What is the area of their intersection (i.e. the shaded region)?

$$circle_1: x^2 + y^2 = 4 \qquad r_1 = 2\\ circle_2: (x-2)^2 + (y-2)^2 = 1 \qquad r_2 = 1\\$$

intersections $$S_1 \text{ and } S_2$$:

$$2 x_{s}^2 - 2a \cdot x_s + a^2 - r_1^2 = 0 \qquad a = \frac{3r_1^2-r_2^2}{2r_1} = \frac{11}{4}\\ \vec{S_1} =\dbinom{ \frac18(11-\sqrt{7}) }{ \frac18(11+\sqrt{7}) }\\ \vec{S_2} =\dbinom{ \frac18(11+\sqrt{7}) }{ \frac18(11-\sqrt{7}) }$$

sectors angels (cosinus-rule) $$\alpha_1 \text{ and } \alpha_2$$:

$$\cos{(\alpha_1)} = \frac{57}{64} \qquad \alpha_1 = 27.0481105464^{\circ}\\ \cos{(\alpha_2)} = \frac{9}{16} \qquad \alpha_2 = 55.7711336722^{\circ}\\$$

Areas sectors $$A_{s_1} \text{ and } A_{s_2}$$:

$$A_{s_1} = 4\pi \frac{27.0481105464^{\circ}}{360^{\circ}}\\ A_{s_2} = \pi \frac{55.7711336722^{\circ}}{360^{\circ}}\\$$

Areas triangles $$A_{t_1} \text{ and } A_{t_2}$$:

$$A_{t_1} = \frac{1}{2} \cdot \left| \vec{S_2}\times \vec{S_1} \right| = \frac{11}{32}\sqrt{7}\\ A_{t_2} = \frac{1}{2} \cdot \left| \left[\vec{S_1}-\binom{2}{2}\right] \times \left[\vec{S_2}-\binom{2}{2}\right] \right|= \frac{5}{32}\sqrt{7}$$

$$\text{The area of their intersection }\quad A = A_{s_1}-A_{t_1}+A_{s_2}-A_{t_2}$$

$$A=\frac{\pi}{360}(4\cdot 27.0481105464^{\circ}+55.7711336722^{\circ})-\frac{ \sqrt{7} }{2}\\ A= 1.43085212603-1.32287565553\\ \mathbf{A=0.10797647050}$$

heureka Oct 28, 2015
#7
+101952
0

I just had my decimal point in the wrong place....Alan said mine is [fairly] correct....

Soooooo.....

A BIG GOLD STAR FOR HEUREKA AND ME....!!!!

And for Melody????....A BAG OF SWITCHES....!!!!

Oct 28, 2015
#8
0

Area of the square :  2 x 2  = 4

Area of Big arc = 1/4 pi 2^2 = pi

Area of small arc = 1/4 pi 1^2 = 1/4 pi

Area OUTSIDE of Small arc = 4 - 1/4 pi

Area OUTSIDE of Big arc = 4 - pi

Add the two outside areas and subtract the square area

4-1/4pi + 4 - pi -4 = 4 - 5/4 pi = ,073 Square units

Oct 28, 2015
#9
+28076
0

This is the same as Melody's first answer Guest.  Good try, but unfortunately it isn't correct - see my first reply above to see why.

Oct 28, 2015
#10
0

Guest 8      Ahhhhh......sorry.  I didn't see the other answers before I posted .   I see your explanation of why my answer is WRONG  and it makes perfect sense '2C' it now.......   ha

Oct 28, 2015