Programming Forums > Community Forums > Coder's Corner Lounge

EverLearning · Jul 25th, 2005, 9:24 PM

CHILL...

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

in red is your error, it should be: - 2ax -2yb: (a - b)^2 = a^2 - 2ab + b^2.
do you HAVE to break it down like this? Language constraints? Why can't you use the one above with sqrt()? I have not programmed extensively with Python, so I'm not sure, but I can check in C/C++.

Quote:

Anyways. Yes, even when I change X, it doesn't consider it in the equation for some reason. I just get like 3.003450100000^e032 or something. I'm using Python.

what values for x are you using?

Sane · Jul 25th, 2005, 9:38 PM

Oh my bad, I thought (x-a)(x-a) = x^2 + (x)(-a)^2 + a^2. Fair enough.

And I don't understand your question, I am using your equation. Just from the X values ...oh my...I think I found the problem. :o , the X range was outside the radius, I thought the equation asked for A and B so that X and Y didn't need to be defined according to the position of the circle.

Okay, it should probably work then. Sorry for the bother.

EverLearning · Jul 25th, 2005, 9:41 PM

Feels good to find your own mistakes, know that feeling

cheers

Navid · Jul 26th, 2005, 1:00 AM

wait...when you say "x^2", do you mean "x squared"?

EverLearning · Jul 26th, 2005, 9:42 AM

Quote:

Originally Posted by navnav

wait...when you say "x^2", do you mean "x squared"?

what do you think? does it look like an integral or smth?

Yes. It is one of the standard ways to express squarede entities without LaTex or superscript tags.

Navid · Jul 26th, 2005, 12:44 PM

But then i can't see how x^2 + y^2 will always equal to 1 in the cordinates of a cirlce

EverLearning · Jul 26th, 2005, 1:30 PM

Quote:

Originally Posted by navnav

But then i can't see how x^2 + y^2 will always equal to 1 in the cordinates of a cirlce

It holds true for UNIT circle, used so much in your higschool trigonometry: r = 1 => r^2 = 1 so x^2 + y^2 = 1.
From general formula in post#3, RHS value is r^2.
Circle is defined as "graph" that consists of all points equidistant from the center, i.e. radius is that distance.

Jul 25th, 2005, 9:38 PM	#12
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,886 Rep Power: 5	Oh my bad, I thought (x-a)(x-a) = x^2 + (x)(-a)^2 + a^2. Fair enough. And I don't understand your question, I am using your equation. Just from the X values ...oh my...I think I found the problem. :o , the X range was outside the radius, I thought the equation asked for A and B so that X and Y didn't need to be defined according to the position of the circle. Okay, it should probably work then. Sorry for the bother. __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

Jul 25th, 2005, 9:41 PM	#13
EverLearning Hobbyist Programmer Join Date: May 2005 Location: Indiana Posts: 130 Rep Power: 4	Feels good to find your own mistakes, know that feeling cheers __________________ got math? yumm... "All men by nature desire to know" -Aristotle, Metaphysics

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Jul 26th, 2005, 1:00 AM	#14
Navid Hobbyist Programmer Join Date: Feb 2005 Location: Canada Posts: 187 Rep Power: 4	wait...when you say "x^2", do you mean "x squared"?

Jul 26th, 2005, 12:44 PM	#16
Navid Hobbyist Programmer Join Date: Feb 2005 Location: Canada Posts: 187 Rep Power: 4	But then i can't see how x^2 + y^2 will always equal to 1 in the cordinates of a cirlce

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)

Programming Forums > Community Forums > Coder's Corner Lounge

The Function Of A Circle.