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DarkBASIC Professional Discussion / Finding the radius ?

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HowDo
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Joined: 28th Nov 2002
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Posted: 3rd Jan 2010 16:01
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Hi All

How do I find out the radius for a circle.

I know the center x,y of the circle but the other end is an object which rotates around this center and moves in and out, giving me many x,y positions.

So how do I make this moving x,y give me the radius I need?

would it be take x2,y2 away from x,y?

Dark Physics makes any hot drink go cold.
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Green Gandalf
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Playing: Malevolence:Sword of Ahkranox, Skyrim, Civ6.
Posted: 3rd Jan 2010 16:21
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No.

Assuming the point (x2, y2) is on the circumference of the circle then you just need the distance between the two points, i.e.



or equivalent expression.

You may need to decide whether you are working with integers or floats, of course.
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HowDo
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Posted: 3rd Jan 2010 16:40
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Cheers Green Gandalf ,work perfect, whats the name of the above calculation.

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Green Gandalf
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Posted: 3rd Jan 2010 16:45 Edited at: 3rd Jan 2010 16:46
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Length of the line joining two points, perhaps?

Or

Distance between two points?

Or

Euclidean distance between two points?
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gbark
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Posted: 3rd Jan 2010 17:24 Edited at: 3rd Jan 2010 17:25
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HowDo, it actually comes from the Pythagorean Theorem. In a right triangle: A^2 + B^2 = C^2

In a euclidean (x/y) coordinate system, such as what DBPro uses, you can effectively treat A and B as the X and Y distances between two points, which gives you a right triangle whose hypotenuse representing the distance "C". Using basic algebra to solve for C, and treating A as the distance between two X points and B as the distance between two Y points, you get C = sqrt((x1-x2)^2+(y1-y2)^2)


But I just call it the "distance equation".
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Veron
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Posted: 3rd Jan 2010 17:29
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You can also use:

(x-a)^2+(y-b)^2 = r^2

Where a/b is the centre of the circle, and r is the radius. You can rearrange it to find either r or a/b.

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Sven B
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Posted: 4th Jan 2010 16:43 Edited at: 4th Jan 2010 16:45
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Quote: "HowDo, it actually comes from the Pythagorean Theorem. In a right triangle: A^2 + B^2 = C^2"


Well, I guess it has multiple names.
Actually, in linear algebra the length of a vector is defined as d = sqrt( <a,a> ) (<a,a> is a notation for a dot product).

This definition is valid in all coordinate systems, as it's only using vectors. It's just that in an Euclidean coordinate system this dot product is written as:
<a,b> = ax*bx + ay*by + az*bz
Combined with:
ab = b - a (Chasles-Mobius)
<ab,ab> = <b-a,b-a> = (bx-ax)^2 + (by-ay)^2 + (bz - az)^2

Another way is of course drawing it and using a right triangle.

So basically you can call it what you want. Pythagoras generally refers to the special case of the cosine formula for triangles:
a^2 = b^2 + c^2 - b*c*cos(alpha) where alpha is the opposite angle of side a.
When the triangle is right, this formula changes to
a^2 = b^2 + c^2 as alpha = 90°

Cheers!
Sven B

[edit] I have no idea why I'm telling this...

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