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Graphing a CircleDate: 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
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