[Guest]

A circle of a radius intersects another circle of radius 20 cm at right angles. What is the difference of the areas of the non-overlapping portion?

What is the sum of the areas of the non-overlapping portion?

[Guest]

I think you should rephrase your question. I think i know what you are trying to say, but It isnt very clear....

if the radii intersect at right angles equidistant from the middle to the edge, the overlapping areas would make an ellipse and you would just have to subtract its area from the area of both circles. buuuttt im lazy and dont feel like doing the math...but that could not be what you were looking for either....

[Guest]

Actually it's simple. If the radii intersect at right angles, they have the same radius. Draw a circle on a grid, then a line from the radius straight up and a line from the radius straight to the left. The other circle intersects to that 2 points lie directly at those points of the circles edge. Like this, only this is squares

________

| ___|__

| | | |

|___|___| |

|______|

So this is actually rather easy to figure out. First you need the distance between the centers. Simple distance formula sqrt((x2 -x1)^2 + (y2 - y1)^2) = 20sqrt(2)

Then you need the angle of the line from an intersection point to r1 to r2. Sounds complicated, but trust me, the formula is cos(x) = (20^2 + 20sqrt(2)^2 - 20^2)/(2*20*20sqrt(2)) This conveniently comes to 45 degrees, which is obvious if you look at the setup, but this would be necessary if the circles were more random. Now we know they are at right angles, otherwise we would need the previous angle *2. so now the formula for the angle of the intersection is..

*note you must use radians. 45 deg = pi/4 90deg = pi/2

(1/2)(pi/2)(20^2) - (1/2)(20^2)sin(pi/2) + (1/2)(pi/4)(20^2) - (1/2)(20^2)sin(pi/4) =

129.81cm

Now the easy part, the area of each circle is pi20^2 = 1256.63 x2 = 2513.27

Now since the intersect is part of both circles we need to subtract that value*2

2513.27 - (129.81*2) = 2253.63cm

Wow ASCII pictures look different once posted

Edited September 8, 2008 by EyesOfTheDead

[Guest]

quote name='EyesOfTheDead' date='Sep 8 2008, 05:41 AM' post='70811']

Actually it's simple. If the radii intersect at right angles, they have the same radius. Draw a circle on a grid, then a line from the radius straight up and a line from the radius straight to the left. The other circle intersects to that 2 points lie directly at those points of the circles edge. Like this, only this is squares

________

| ___|__

| | | |

|___|___| |

|______|

So this is actually rather easy to figure out. First you need the distance between the centers. Simple distance formula sqrt((x2 -x1)^2 + (y2 - y1)^2) = 20sqrt(2)

Then you need the angle of the line from an intersection point to r1 to r2. Sounds complicated, but trust me, the formula is cos(x) = (20^2 + 20sqrt(2)^2 - 20^2)/(2*20*20sqrt(2)) This conveniently comes to 45 degrees, which is obvious if you look at the setup, but this would be necessary if the circles were more random. Now we know they are at right angles, otherwise we would need the previous angle *2. so now the formula for the angle of the intersection is..

*note you must use radians. 45 deg = pi/4 90deg = pi/2

(1/2)(pi/2)(20^2) - (1/2)(20^2)sin(pi/2) + (1/2)(pi/4)(20^2) - (1/2)(20^2)sin(pi/4) =

129.81cm

Now the easy part, the area of each circle is pi20^2 = 1256.63 x2 = 2513.27

Now since the intersect is part of both circles we need to subtract that value*2

2513.27 - (129.81*2) = 2253.63cm

Wow ASCII pictures look different once posted

[Guest]

A circle of a radius intersects another circle of radius 20 cm at right angles. What is the difference of the areas of the non-overlapping portion?

What is the sum of the areas of the non-overlapping portion?

I'm not exactly sure about the phrasing of the OP, but

There are a couple ways two circles can intersect each other at right angles. One way is if the circles lie in different, perpendicular planes. In that case, the area of the overlap would be zero.

If they are coplanar, then the question is ambiguous because it does not specify the radius of the first circle. The most obvious way is if they have the same radius (like EotD found), but that doesn't have to be the case.

[Guest]

I have to confess i'm not sure of the answer but i do know that the lines of the circle cross over eac other so the right angle, inreality imaginry ones, are foarmed.

hope that helps.

[Guest]

I have to confess i'm not sure of the answer but i do know that the lines of the circle cross over eac other so the right angle, inreality imaginry ones, are foarmed.

hope that helps.

how does 2 circles on the same plane intersect each other at 90 degrees ??? Where does the 90 degree form ?

[HoustonHokie]

I didn't work out all the fancy math - I just drew the thing in CADD and measured areas. Assuming equal radii and that the first part of the problem is looking for the purple area and the second part is looking for red plus blue, the numbers are 228 cm² and 2056 cm². Note: the overlap is not an ellipse - it's more like a football.

Some have asked where the right angle intersection occurs. It occurs in two locations in this picture. The instantaneous angle of any location on a circle is perpedicular to the radius to that point. So the two locations where the red and blue circles intersect are perpendicular to each other, because their radials are perpendicular to each other.

circles.bmp

[Guest]

If they are coplanar, then the question is ambiguous because it does not specify the radius of the first circle. The most obvious way is if they have the same radius (like EotD found), but that doesn't have to be the case.

The only solution is for both circles to have the same radius (unless you allow zero and infinite radius).

[Prof. Templeton]

The radius of each cirle is 20, so the area of each is pi*R² = 1256.77. Now imagine a square drawn around where they intersect, such a square would be 20 by 20 with an area of 400 and cut off 1 quarter of each circle. Draw a diagonal through the square halving it which will bisect our "football". 1 quarter of the area of 1 circle is 314.92 (hmmm 100 x pi) minus 200 and thats the amount 1 circle sticks over the diagonal and multiply by 2 for the whole "football". 228.38 cm².

For the second part each circle is missing 228.38 so it's (1256.77 - 228.38) * 2 or 2056.78.

[Guest]

The only solution is for both circles to have the same radius (unless you allow zero and infinite radius).

Oops. Made a mistake in my calculation. The radius of the second circle can be anything...

[Guest]

how does 2 circles on the same plane intersect each other at 90 degrees ??? Where does the 90 degree form ?

For 2 circles to intersect they need to do so at 2 points. You draw a line from one intersection point, to the radius, to the other intersection point.

[Prof. Templeton]

Oops. Made a mistake in my calculation. The radius of the second circle can be anything...

That would only create 1 right angle. The OP says angles.

[Guest]

That would only create 1 right angle. The OP says angles.

I think you guys are reading too far into it.

my understanding would make the radii intersect at each halfway point which would be 10...

so far my answer lies with Eyes...although i haven't double checked the math.

[Guest]

The only solution is for both circles to have the same radius (unless you allow zero and infinite radius).

At first I thought that wasn't true, because I saw the infinite radius solution and thought there must be solutions in between. I tried to solve it algebraically, but that got really complicated really quickly.

So I went through and found a geometric proof that is very convincing. You're right, it is not ambiguous in the coplanar case. Live and learn.

[Guest]

That would only create 1 right angle. The OP says angles.

Not true. See example attached. Both intersections are perpendicular. (edit: well, they're supposed to be anyway... I can't draw)

The requirement is for \alpha + \beta = pi. Then, because the quadrilateral can be bisected into two similar triangles, the two remaining angles are both pi/2.

If the radius of the large circle is R and of the smaller circle is r, then:

R/r = tan(alpha/2)

Edited September 8, 2008 by d3k3

[Guest]

At first I thought that wasn't true, because I saw the infinite radius solution and thought there must be solutions in between. I tried to solve it algebraically, but that got really complicated really quickly.

So I went through and found a geometric proof that is very convincing. You're right, it is not ambiguous in the coplanar case. Live and learn.

Alright, I'm flip-flopping on this one. I am now convinced that you can have two right angles with any two radii.

EDIT: Removed diagram I was working on to prove the radii had to be equal. Forgot to remove the attachment when I realized I was wrong.

EDIT: I agree with d3k3, this time for real.

Edited September 8, 2008 by Chuck Rampart

[Prof. Templeton]

Not true. See example attached. Both intersections are perpendicular. (edit: well, they're supposed to be anyway... I can't draw)

The requirement is for \alpha + \beta = pi. Then, because the quadrilateral can be bisected into two similar triangles, the two remaining angles are both pi/2.

If the radius of the large circle is R and of the smaller circle is r, then:

R/r = tan(alpha/2)

I stand corrected (well I'm sitting anyway).

[Guest]

A circle of a radius intersects another circle of radius 20 cm at right angles. What is the difference of the areas of the non-overlapping portion?

What is the sum of the areas of the non-overlapping portion?

I feel like its unclear what perpendicular means here. My first thought was that the tangent lines at the points where the circles intersect are perpendicular. But that would allow for a circle of any size.

[Guest]

I think you guys are reading too far into it.

my understanding would make the radii intersect at each halfway point which would be 10...

so far my answer lies with Eyes...although i haven't double checked the math.

Thanks Raven. I am taking a simulation course in collision detection so this came second hand to me. Assuming I myself assumed everything correctly, this is the formula to calculate the intersection of 2 circles.

[Prime]

It seems to me, the only way to disambiguate this problem, is to use Chuck's suggestion that the circles lie in perpendicular planes.

Otherwise, draw a circle with radius 20. Draw any perpendicular line segment of any length intersecting a radius of that circle, use it as a radii for another circle.

Or draw a tangent line to the first circle, use that as a radius for another circle. In such ways two circles may intersect whichever way, or one could be enclosed entirely inside another, or they could be completely outside each other and you can extend their radii to intersect and show that they are at the right angle to each other.