Equation of a CircleDate: 05/13/2003 at 13:52:22 From: Balbino Subject: Equation of a circle Find the equation of a circle with the center at point (3,-4) and radius 2. Date: 05/14/2003 at 14:23:04 From: Doctor Dotty Subject: Re: Equation of a circle Hi Balbino, Thanks for the question. Here's a circle with centre (0,0) and radius R: | | ,---+---. ,-' y+ `-P(x,y) ,' | / `. ,' | / `. / | R/ \ ; | / : ; | / : ; |/ : -------+-------------+------+------+------ : | x ; : | ; : | ; \ | / `. | ,' `. | ,' `-. | ,-' `---+---' | | Pythagoras' Theorem tells us that R^2 = x^2 + y^2. If you think about it, that equation is true whatever x and y are. If they are in a negative quadrant, the radius is still positive. Can you see why? So, that's a circle with centre (0,0). We don't want this; we want it at another centre. So what if the centre is at (a,b)? | y+ ,-----.P(x,y) | ,' /`. | / R/ \ | ; / : b+ | + | | : (a,b) ; | \ / | `. ,' | '-----' -----+--------+--+-------- | a x | Now, with Pythagoras' Theorem again: R^2 = (the distance from x to a)^2 + (the distance from y to b)^2 R^2 = (x - a)^2 + (y - b)^2 That is the equation of a circle. So for a circle of centre (-10,5) and radius 3, we get: 3^2 = (x - -10)^2 + (y - 5)^2 9 = (x + 10)^2 + (y - 5)^2 Some people prefer a general equation where the brackets have been multiplied out, so that it is in the form: x^2 + y^2 + 2gx + 2fy + c = 0 In our case, that can be worked out: 9 = (x + 10)^2 + (y - 5)^2 9 = x^2 + 20x + 100 + y^2 - 10y + 25 9 = x^2 + y^2 + 20x - 10y + 125 0 = x^2 + y^2 + 20x - 10y + 116 Can you now find the equation of your circle? Write back if I can be of any more help - on this or anything else. - Doctor Dotty, The Math Forum http://mathforum.org/dr.math/ |
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