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 Algebra Pre-Algebra and Basic Algebra Math Forum

 December 7th, 2012, 07:28 AM #1 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 circle [color=#008000]Dear All! If we have: http://www.mathsisfun.com/algebra/circle-equations.html does anyone have the good pic so that we can be sure that we MIGHT use a=1, b=2 and c=3? Thanks! [/color]
 December 7th, 2012, 08:29 AM #2 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: circle [color=#0000FF]Well, I think we can choose a, b and c as we want, but we CAN'T choose A, B and C, this HAS to be calculated...[/color]
 December 7th, 2012, 08:43 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: circle a=1, b=2, c=3 CANNOT be a triangle. What is your question?
 December 7th, 2012, 01:24 PM #4 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: circle [color=#008000](x-a)^2+(y-b)^2=c^2, can be a triangle. x^2+y^2+Ax+By+c=0 has just A$\not=$a, B$\not=$b and C$\not=$c Have a look at the pic I linked! Sides of the triangle are not a, b, c but (x-a), (y-b), c.[/color]
 December 7th, 2012, 03:21 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 As (a, b) specifies where the circle is, a and b can have any values. It's clear that c = r, so c can have any positive value. If you had c = 0, the "circle" would be just the single point (a, b). The linked site already has good pictures.

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