Graphing a Circle
Date: 04/14/98 at 15:02:55 From: Philip J. Peisch Subject: algebra I would like to know how to graph the equation of a circle. I have a big test coming up - can you please help me? Thanks - Philip J. Peisch
Date: 04/14/98 at 15:48:57 From: Doctor Kate Subject: Re: algebra The equation of a circle is usually: (x-a)^2 + (y-b)^2 = r^2 (Here "^2" means "squared"). If your equation isn't in that form, you will have to use algebraic techniques to get it into that form. I won't go into that in too much depth if you don't need it, but here's an example: x^2 + y^2 + 2y + 1 = 36 You've probably done some factoring, so you'll notice that y^2 + 2y + 1 is just (y+1)^2, and also that 36 = 6^2. So your equation is then: (x-0)^2 + [y-(-1)]^2 = 6^2 (here a = 0, b = -1 and r = 6) Let's go back to (x-a)^2 + (y-b)^2 = r^2 To graph this, you notice two things: - The origin (centre) of your circle is at (a,b). (Notice that in the example I gave above, b is -1 not 1. Don't make that mistake on your test). - The radius of your circle is r. So draw your circle with that information. Pick the centre and then pick some points that are r units away from it and draw the circle through those (so you might draw (a, b+r), (a, b-r), (a+r, b) and (a-r, b) as four points and draw a circle that passes through all four). So in the example above, the centre of your circle is (0, -1) and it has a radius (not diameter) of 6 units. So it will pass through the points (0, 5), (0, -7), (6, -1) and (-6, -1). By the way, why is the centre (a,b) and the radius r? Well in x^2 + y^2 = r^2, it's just the Pythagorean Theorem. And then going to (x-a) and (y-b) is just a translation of your graph. These are interesting ideas; please think about them (they are more important things to know than just 'how to graph a circle'). If you have more questions, please write us, we're happy to help. Hope this helps and good luck on your test. -Doctor Kate, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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