Re-asking a question that I made 2 years ago and couldn't get to an answer. Maybe now someone can reach it!
Two circles, C and Cr intersect each other. C's radius is 1, while Cr's radius is "r". Their tangent lines form an internal angle of 120° at each intersection point. 1) Calculate the distance "d", which is the distance between the centers of both circles, in terms of r. 2) Find where d has a maximum or a minimum. 3) Find the area of the intersection in terms of π

- ChillOut

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- ChillOut

|dw:1437616135891:dw|

- ChillOut

|dw:1437616351195:dw|

- ChillOut

I'm starting to think that this is unsolvable.

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## More answers

- ChillOut

All right, so I found this: http://www.had2know.com/academics/intersection-angle-two-circles.html Is this of any help?

- ChillOut

I swear if I get an answer to this I'll become a math teacher.

- ganeshie8

Cutting out 30 degrees from either ends of the tangent lines, it is easy to see that the angle made by radii segments at the intersection point is 60 degrees :
|dw:1437618536980:dw|

- ChillOut

How do I prove that it's 60°?

- ChillOut

If it's right, I can just go on with implicit differentiation, yeah?

- ganeshie8

Easy : draw a better diagram that actually represents 120 degrees between the tangents, then argue that the perpendicular to tangent line at the point of tangency passes through the center

- ChillOut

Oh, yeah. I see it now. Now I think I can finish it.

- ChillOut

I was failing to notice how the radii and the tangents connected to form a triangle.

- ganeshie8

We don't really need calculus for second part, we may use the fact that \(d\) is positive and the max/min value of \(d\) occurs at the same point as \(d^2\)

- ganeshie8

|dw:1437619117612:dw|

- ChillOut

Yeah, I did the whole d'(r) and found 1/2 as well.

- ganeshie8

*min value for \(d\) occurs at \(r = 1/2\)

- jim_thompson5910

I'm getting d^2 = 1 + r^2 + r

- ChillOut

Just for future reference, leaving the derivative here: \(\frac{dd}{dr}=(2r-1)*(2(r²-r+1)^{0.5}\))

- jim_thompson5910

Law of Cosines
c^2 = a^2 + b^2 - 2*a*b*cos(C)
d^2 = 1^2 + r^2 - 2*1*r*cos(120)
d^2 = 1^2 + r^2 - 2*1*r*(-0.5)
d^2 = 1^2 + r^2 + r

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- ChillOut

@jim_thompson5910 , why is the angle between the radii 120º in your picture?

- ganeshie8

@jim_thompson5910 looks 120 degrees has to be the other angle :
|dw:1437619755534:dw|

- ChillOut

Now I am confused, If they are orthogonal to the tangent lines, they should have the same angle between them, right?

- jim_thompson5910

look at the attached pic I'm posting
angle CAF = 90 degrees since the circle is tangent at point A
angle GAD = 90 degrees (same reason)
angle DAF = 30 degrees because it's the solution so 90+x = 120
angle DAC = (angle DAF) + (angle CAF)
angle DAC = (30) + (90)
angle DAC = 120

##### 1 Attachment

- jim_thompson5910

|dw:1437619928573:dw|

- ganeshie8

again, the construction is incorrect as 120 degrees has to be the internal angle as per the question

- jim_thompson5910

|dw:1437619979044:dw|

- jim_thompson5910

oh did I flip the angles?

- ganeshie8

I think so, after flipping the angles, ur construction also gives the same expression
\(d^2 = 1+r^2-r\)

- ChillOut

|dw:1437620217558:dw| This one is supposed to have 120°

- ChillOut

So, by using the law of cosines and assuming it's 60° it gives @ganeshie8's expression

- ganeshie8

It is easy to prove using that new better picture, lets do it

- ganeshie8

|dw:1437620436117:dw|

- ganeshie8

|dw:1437620527012:dw|

- ganeshie8

Angle between radius and tangent = 90
therefore angle between radius and other circle's tangent = 120-90 = 30
similar reasoning yields the other angle also 30

- ganeshie8

|dw:1437620634836:dw|

- ganeshie8

so the angle between radii segments = 120-30-30 = 60

- ChillOut

Yes, pretty nice! We already have the critical point for the radii. Now I need to remember how to answer c) without resorting to calculus

- ChillOut

Just find the area of both sectors segments and add them?

- ganeshie8

calculus is always there if we forget geometry :)
lets see if this can be done smartly w/o calc

- ChillOut

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- ChillOut

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- ChillOut

thing is... I need to find Beta.

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- ChillOut

Or do I?

- ChillOut

Well, I remember the area of a circle's sector. It's given by \(A=r²*arccos(d/2r)\) Where d is the distance between the centers and r the radius of the circle in question.

- ganeshie8

you want to find the area for general case, in terms of \(r\), right ?

- ChillOut

This should be ideal.

- ChillOut

Well, I'm a lackluster in geometry, as you can see.

- ganeshie8

your method works, start by finding this central angle, \(x\) :
|dw:1437623030778:dw|

- ganeshie8

once you have \(x\), you can find the area of sector and subtract the area of triangle from it to get :
|dw:1437623216203:dw|

- ganeshie8

you can do the same for the other piece also! looks good!

- ChillOut

The line that cuts through both intersections also divides the angles equally, right?

- ganeshie8

Nope.

- ganeshie8

use base angles of isosceles triangle theorem

- ganeshie8

|dw:1437623442647:dw|

- ganeshie8

applying cosine law in that triangle with sides \(1, r, d\) :
\[r^2 = 1^2+d^2-2d\cos(x/2)\]
you can find \(x\)

- ChillOut

Wouldn't I be able to solve for x only i I had a particular solution for r, which isn't the case?

- ganeshie8

you just want to find the overlapping area for \(r=1/2\) is it ?

- ChillOut

Yeah. I don't see any way to find a general solution as d is defined by r.

- ganeshie8

applying cosine law in that triangle with sides \(1, r, d\) :
\[r^2 = 1^2+d^2-2d\cos(x/2)\]
\[r^2 = 1+(1+r^2-r)-2d\cos(x/2) \implies x = 2\arccos\left(\dfrac{2-r}{2d}\right)\]

- ganeshie8

\(d\) is just a number
as you can see, we can always replace \(d^2\) by \(1+r^2-r\)

- ChillOut

All right, I need to calculate the sector's area.

- ChillOut

\((x/2\pi)*\pi*r²\)

- ganeshie8

Area of shaded region in circle of radius \(1\) :
\[\pi*\dfrac{x}{2\pi} = \arccos\left(\dfrac{2-r}{2d}\right)\]

- ChillOut

Just subtract the triangle area and we have the segment, right?

- ganeshie8

Yes, just algebra..

- ChillOut

So, \(A_{segment}=arccos(\frac{2-r}{2d})-\frac{1*1*sin(x)}{2}\)

- ChillOut

which in turn is \(arccos(\frac{2-r}{2d})-sin(2arccos(\frac{2-r}{2d})) \)

- ChillOut

Oops! Missed /2 in the second part.

- ganeshie8

Looks good to me!
Area of shaded region in circle of radius \(1\) :
\(\large \arccos(\dfrac{2-r}{2d})-\dfrac{\sin(2\arccos(\frac{2-r}{2d}))}{2} \)
|dw:1437625381250:dw|

- ChillOut

Now for the other sector... Which will be in terms of "r"

- ganeshie8

|dw:1437625666331:dw|

- ganeshie8

do we get
\[y=2\arccos\left(\dfrac{2r^2-r }{ 2dr}\right)\]
?

- ChillOut

Yes.

- ChillOut

Wait... Let me take a second look.

- ChillOut

Yeah, that's what I got.

- ChillOut

\(A_{segment}=r^{2}arccos(\frac{2r^{2}-r}{2dr})-\frac{r^{2}sin(2arccos(\frac{2r^{2}-r}{2dr}))}{2}\)

- ChillOut

For the circle of radius r. Sorry for the small TeX text

- ChillOut

Then, adding the segments' areas, we have \(arccos(\frac{2-r}{2d})-\frac{sin(arccos(\frac{2-r}{2d})}{2}+r²arccos(\frac{2r²-r}{2dr})-\frac{r²sin(2arccos(\frac{2r²-r}{2dr}))}{2}\)

- ChillOut

I guess this is it!

- ChillOut

P.S: Missed a "2" before the first arc cosine.

- ganeshie8

Looks neat!

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