ChillOut
  • ChillOut
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ChillOut
  • ChillOut
|dw:1437616135891:dw|
ChillOut
  • ChillOut
|dw:1437616351195:dw|
ChillOut
  • ChillOut
I'm starting to think that this is unsolvable.

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ChillOut
  • ChillOut
All right, so I found this: http://www.had2know.com/academics/intersection-angle-two-circles.html Is this of any help?
ChillOut
  • ChillOut
I swear if I get an answer to this I'll become a math teacher.
ganeshie8
  • 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
  • ChillOut
How do I prove that it's 60°?
ChillOut
  • ChillOut
If it's right, I can just go on with implicit differentiation, yeah?
ganeshie8
  • 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
  • ChillOut
Oh, yeah. I see it now. Now I think I can finish it.
ChillOut
  • ChillOut
I was failing to notice how the radii and the tangents connected to form a triangle.
ganeshie8
  • 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
  • ganeshie8
|dw:1437619117612:dw|
ChillOut
  • ChillOut
Yeah, I did the whole d'(r) and found 1/2 as well.
ganeshie8
  • ganeshie8
*min value for \(d\) occurs at \(r = 1/2\)
jim_thompson5910
  • jim_thompson5910
I'm getting d^2 = 1 + r^2 + r
ChillOut
  • ChillOut
Just for future reference, leaving the derivative here: \(\frac{dd}{dr}=(2r-1)*(2(r²-r+1)^{0.5}\))
jim_thompson5910
  • 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
ChillOut
  • ChillOut
@jim_thompson5910 , why is the angle between the radii 120º in your picture?
ganeshie8
  • ganeshie8
@jim_thompson5910 looks 120 degrees has to be the other angle : |dw:1437619755534:dw|
ChillOut
  • ChillOut
Now I am confused, If they are orthogonal to the tangent lines, they should have the same angle between them, right?
jim_thompson5910
  • 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
jim_thompson5910
  • jim_thompson5910
|dw:1437619928573:dw|
ganeshie8
  • ganeshie8
again, the construction is incorrect as 120 degrees has to be the internal angle as per the question
jim_thompson5910
  • jim_thompson5910
|dw:1437619979044:dw|
jim_thompson5910
  • jim_thompson5910
oh did I flip the angles?
ganeshie8
  • ganeshie8
I think so, after flipping the angles, ur construction also gives the same expression \(d^2 = 1+r^2-r\)
ChillOut
  • ChillOut
|dw:1437620217558:dw| This one is supposed to have 120°
ChillOut
  • ChillOut
So, by using the law of cosines and assuming it's 60° it gives @ganeshie8's expression
ganeshie8
  • ganeshie8
It is easy to prove using that new better picture, lets do it
ganeshie8
  • ganeshie8
|dw:1437620436117:dw|
ganeshie8
  • ganeshie8
|dw:1437620527012:dw|
ganeshie8
  • 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
  • ganeshie8
|dw:1437620634836:dw|
ganeshie8
  • ganeshie8
so the angle between radii segments = 120-30-30 = 60
ChillOut
  • 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
  • ChillOut
Just find the area of both sectors segments and add them?
ganeshie8
  • ganeshie8
calculus is always there if we forget geometry :) lets see if this can be done smartly w/o calc
ChillOut
  • ChillOut
1 Attachment
ChillOut
  • ChillOut
1 Attachment
ChillOut
  • ChillOut
thing is... I need to find Beta.
1 Attachment
ChillOut
  • ChillOut
Or do I?
ChillOut
  • 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
  • ganeshie8
you want to find the area for general case, in terms of \(r\), right ?
ChillOut
  • ChillOut
This should be ideal.
ChillOut
  • ChillOut
Well, I'm a lackluster in geometry, as you can see.
ganeshie8
  • ganeshie8
your method works, start by finding this central angle, \(x\) : |dw:1437623030778:dw|
ganeshie8
  • 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
  • ganeshie8
you can do the same for the other piece also! looks good!
ChillOut
  • ChillOut
The line that cuts through both intersections also divides the angles equally, right?
ganeshie8
  • ganeshie8
Nope.
ganeshie8
  • ganeshie8
use base angles of isosceles triangle theorem
ganeshie8
  • ganeshie8
|dw:1437623442647:dw|
ganeshie8
  • 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
  • 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
  • ganeshie8
you just want to find the overlapping area for \(r=1/2\) is it ?
ChillOut
  • ChillOut
Yeah. I don't see any way to find a general solution as d is defined by r.
ganeshie8
  • 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
  • ganeshie8
\(d\) is just a number as you can see, we can always replace \(d^2\) by \(1+r^2-r\)
ChillOut
  • ChillOut
All right, I need to calculate the sector's area.
ChillOut
  • ChillOut
\((x/2\pi)*\pi*r²\)
ganeshie8
  • ganeshie8
Area of shaded region in circle of radius \(1\) : \[\pi*\dfrac{x}{2\pi} = \arccos\left(\dfrac{2-r}{2d}\right)\]
ChillOut
  • ChillOut
Just subtract the triangle area and we have the segment, right?
ganeshie8
  • ganeshie8
Yes, just algebra..
ChillOut
  • ChillOut
So, \(A_{segment}=arccos(\frac{2-r}{2d})-\frac{1*1*sin(x)}{2}\)
ChillOut
  • ChillOut
which in turn is \(arccos(\frac{2-r}{2d})-sin(2arccos(\frac{2-r}{2d})) \)
ChillOut
  • ChillOut
Oops! Missed /2 in the second part.
ganeshie8
  • 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
  • ChillOut
Now for the other sector... Which will be in terms of "r"
ganeshie8
  • ganeshie8
|dw:1437625666331:dw|
ganeshie8
  • ganeshie8
do we get \[y=2\arccos\left(\dfrac{2r^2-r }{ 2dr}\right)\] ?
ChillOut
  • ChillOut
Yes.
ChillOut
  • ChillOut
Wait... Let me take a second look.
ChillOut
  • ChillOut
Yeah, that's what I got.
ChillOut
  • ChillOut
\(A_{segment}=r^{2}arccos(\frac{2r^{2}-r}{2dr})-\frac{r^{2}sin(2arccos(\frac{2r^{2}-r}{2dr}))}{2}\)
ChillOut
  • ChillOut
For the circle of radius r. Sorry for the small TeX text
ChillOut
  • 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
  • ChillOut
I guess this is it!
ChillOut
  • ChillOut
P.S: Missed a "2" before the first arc cosine.
ganeshie8
  • ganeshie8
Looks neat!

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