Area of an arc

[BMAD]

Suppose that there is an arc from a circle going through the point (0,3) and (2,0).  

How many different circles (different radius, different centers) could make this arc?

Without using calculus or trig, find the area under the arc between the x and y axis? 

Edited June 1, 2016 by BMAD

[CaptainEd]

Interesting puzzle, BMAD. i'm falling for it. I get the feeling I'm missing a key ingredient or inference. at any rate, my naive views:

Spoiler

number of circles

infinite: any center on the line y=3*x/2-1.25

Spoiler

area under the arc

it depends: if the center is at the center of the Sun, on the line given above, at x~( - 93M miles ), then the area under the arc is essentially the area of the triangle. But if the center is at x ~ ( 93M miles ), then the area under the arc is essentially 0, and if the center is at ( 1.5, 1 ), then it's more than those, but I haven't figured out how much...

 

here's how much: 

Spoiler

for circle at ( 1.5, 1 )
r = sqrt (9/4 + 1) = sqrt ( 13/4 )
area of circle is pi*13/4
area under arc is half the circle plus the triangle
= pi * 13/8 + 3

 

Edited June 1, 2016 by CaptainEd

[BMAD]

9 hours ago, CaptainEd said:

Interesting puzzle, BMAD. i'm falling for it. I get the feeling I'm missing a key ingredient or inference. at any rate, my naive views:

  Hide contents

number of circles

infinite: any center on the line y=3*x/2-1.25

  Reveal hidden contents

area under the arc

it depends: if the center is at the center of the Sun, on the line given above, at x~( - 93M miles ), then the area under the arc is essentially the area of the triangle. But if the center is at x ~ ( 93M miles ), then the area under the arc is essentially 0, and if the center is at ( 1.5, 1 ), then it's more than those, but I haven't figured out how much...

 

here's how much: 

  Hide contents

for circle at ( 1.5, 1 )
r = sqrt (9/4 + 1) = sqrt ( 13/4 )
area of circle is pi*13/4
area under arc is half the circle plus the triangle
= pi * 13/8 + 3

 

I am curious by how you found your line and how you know the line works?  

Is there another possible line containing solutions?

9 hours ago, CaptainEd said:

Interesting puzzle, BMAD. i'm falling for it. I get the feeling I'm missing a key ingredient or inference. at any rate, my naive views:

  Reveal hidden contents

number of circles

infinite: any center on the line y=3*x/2-1.25

  Reveal hidden contents

area under the arc

it depends: if the center is at the center of the Sun, on the line given above, at x~( - 93M miles ), then the area under the arc is essentially the area of the triangle. But if the center is at x ~ ( 93M miles ), then the area under the arc is essentially 0, and if the center is at ( 1.5, 1 ), then it's more than those, but I haven't figured out how much...

 

here's how much: 

  Reveal hidden contents

for circle at ( 1.5, 1 )
r = sqrt (9/4 + 1) = sqrt ( 13/4 )
area of circle is pi*13/4
area under arc is half the circle plus the triangle
= pi * 13/8 + 3

 

Though I don't disagree with your assertion that there are infinite circles I do not believe the domain of your line is correct. 

Edited June 2, 2016 by BMAD

[BMAD]

22 hours ago, CaptainEd said:

Interesting puzzle, BMAD. i'm falling for it. I get the feeling I'm missing a key ingredient or inference. at any rate, my naive views:

  Reveal hidden contents

number of circles

infinite: any center on the line y=3*x/2-1.25

  Hide contents

area under the arc

it depends: if the center is at the center of the Sun, on the line given above, at x~( - 93M miles ), then the area under the arc is essentially the area of the triangle. But if the center is at x ~ ( 93M miles ), then the area under the arc is essentially 0, and if the center is at ( 1.5, 1 ), then it's more than those, but I haven't figured out how much...

 

here's how much: 

  Reveal hidden contents

for circle at ( 1.5, 1 )
r = sqrt (9/4 + 1) = sqrt ( 13/4 )
area of circle is pi*13/4
area under arc is half the circle plus the triangle
= pi * 13/8 + 3

 

after testing your center point of (1.5,1), I have found that it doesn't work in producing a consistent radius.  You are using your formula incorrectly i think.

[CaptainEd]

Oops,

Spoiler

 

I reversed the coordinates!

Line connecting the points is y = - 3x  / 2 + 3

Perpendicular bisector, where all the centers lie is y = 2x / 3 + 5/6

Center of smallest circle, directly between the two endpoints of the arc is ( 1, 1.5 )

area is still right, I think.

 

 

[Buddyboy3000]

The line that I got was the same as CaptainEd got. However, this is what I got for finding the circles...

Spoiler

(x - 1 - a)² + (y - 1.5 - 2a / 3) = ((a / 1.5)² + 1) * 3.25

This is the equation for all the circles that will pass through the points (0,3) and (2,0), which is an infinite amount.

The x and y will fall on the line 2x / 3 +5/6, which was given by CaptainEd. The value of A is a variable to move the circle position and radius. If A is 0, then the circle will be the smallest one possible with a center exactly between the two points. As A gets larger in absolute value, then circle will get larger too, with a center point farther away. This is to prove that the amount of circles that can go through the two points are infinite.

I have not yet figured out the second part of the problem with the area, but this is what I have for now.

[BMAD]

2 hours ago, Buddyboy3000 said:

The line that I got was the same as CaptainEd got. However, this is what I got for finding the circles...

  Hide contents

(x - 1 - a)² + (y - 1.5 - 2a / 3) = ((a / 1.5)² + 1) * 3.25

This is the equation for all the circles that will pass through the points (0,3) and (2,0), which is an infinite amount.

The x and y will fall on the line 2x / 3 +5/6, which was given by CaptainEd. The value of A is a variable to move the circle position and radius. If A is 0, then the circle will be the smallest one possible with a center exactly between the two points. As A gets larger in absolute value, then circle will get larger too, with a center point farther away. This is to prove that the amount of circles that can go through the two points are infinite.

I have not yet figured out the second part of the problem with the area, but this is what I have for now.

Does the area approach anything as A goes to infinity?

Also, I still question the domain of this line. I don't believe a value is feasible for all real numbers 

 

[CaptainEd]

@BMAD, I've interpreted the goal as being to give the area meeting ALL of these constraints: ( a ) under the arc, ( b ) within the body of the circle, ( c ) above x-axis and to the right of the y-axis.

Once the center of the circle moves to the right and up along the line I gave, this area shrinks until it is a thin lens connecting (0,3) and (2,0), with a limit of 0. Have I still got the formula wrong? Or is your question only about the domain of x?

[Buddyboy3000]

The line is the area that all the centers can be at. No center of a circle that goes through the two points can be off that line.

1. All points off a circle have to be the same distance to the center, which is the radius. This goes also with points (0,3) and (2,0). Imagine a line connecting them to the center of the circle. These lines would be marked as the variable X. So, you form a isosceles triangle. Two sides are X, and the other is the sqrt(13), which is the line connecting the two points to each other. In an isosceles triangle the top would be directly above the center of the base. In this case, the base would be the sqrt(13). The top would be the center of the circle. This is where you would get the line, because it is perpendicular with the line connecting the two points, and intersects through the middle of it.

2. To answer the first thing you asked me, the variable A will affect the area. It depends on which arc you want to measure, the minor arc or the major arc. As A reaches towards positive infinity, the area with the minor arc will get bigger, and the same thing with the major arc. When A goes the opposite way, the minor arc will get smaller in area, and the major arc will forever be at 0. Bigger and smaller area is based on when A is 0, and the circle is at its smallest. When I say A gets larger in absolute value, this means that the circle will just get bigger.

This is why I do not believe that there can be a center anywhere off this line. If so, then your question has me stumped. 

Edited June 3, 2016 by Buddyboy3000
Hit the post button too early

[CaptainEd]

@BMAD, I see I am being inconsistent.

If the center is on the line that I gave, and x <= 1, the arc is at the top of the circle, and above the line from (0,3) to (2,0).

If the center is exactly at (1,1.5), then the area is as I said above.

if x > 1, then there is an arc of the bottom of the circle in the quadrant (x>0, y>0). Then the area under the arc is OUTSIDE the circle, so we'd have to know just what area you are interested in, in that case.

Edited June 3, 2016 by CaptainEd

[BMAD]

@CaptainEd

Forgive if my English is not clear this happens to me  a lot... 

But unless my intuition is wrong... 

If the center of the circle is southwest of the midpoint between the two values (1,1.5) then the arc curves away from the origin.  If the center of the circle is northeast of the midpoint then the arc curves towards the origin while intersecting the two points.  Then it should follow that there is a point, somewhere on that lone where the arc curves towards and away from the origin while going through the point at the same time (making the axis tangent I believe) or it neither curves towards or away from the origin. 

The last case,  I argue does not meet the conditions of the question but maybe I over thought this one. 

@Buddyboy3000 I used an approach similar to your similar triangles in conjunction with Captain Ed's formula

[bonanova]

 

Spoiler

First off, a circle is determined uniquely by three distinct points on its circumference. If the points are collinear, the circle has infinite radius. We are given two circumference points, (0,3) and (2,0) and are asked, not about other points on the circumference but about possible locations of the center point. Clearly the center must lie along the perpendicular bisector of the segment defined by the given points (gray line segment), so that its distance to them is the same. That (red) line has slope 2/3 (negative reciprocal of -3/2) a y-intercept of 5/6 and an x-intercept of -5/4.

When the center's x-coordinate (along the red line) is 0 or 2, the circles are tangent to the x-axis; when its y-coordinate is 0 or 3, they are tangent to the y-axis. The general behavior of the circles as the center moves along the red line is seen by these four instances. I'm not sure how to calculate the area subtended by the arc between the two given points and the positive x- and y- axes without recourse to calc or trig, even if the area of a circular sector is known: in some cases the caps intersect.I thought first of rotating the drawing so that the red and gray lines align with the principal axes, but that does not seem to simplify the calculation.

 

 

[bonanova]

I don't know how to do  without calc or trig ... interesting puzzle