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 August 12th, 2012, 02:17 PM #1 Newbie   Joined: Aug 2012 Posts: 5 Thanks: 0 The Radius of a Circle Given this equation: x^2 + y^2 - 2x - 2y = -9 Is this the equation of a circle? It appears to me that this circle has a negative radius, and this does not make sense to me. The square root of -9 is imaginary, and therefore it seems that this equation cannot graph a circle.
August 12th, 2012, 04:04 PM   #2
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Re: The Radius of a Circle

Hello, awinston!

Quote:
 Given this equation: $x^2\,+\,y^2\,-\,2x\,-\,2y\:=\:-9$ Is this the equation of a circle? It appears to me that this circle has a negative radius, and this does not make sense to me. The square root of -9 is imaginary, and therefore it seems that this equation cannot graph a circle.

You happen to be correct . . . but for the wrong reason.

The minus on the "9" is not the proper indicator.

We must complete the square first.

$\text{W\!e have: }\:x^2\,-\,2x\,+\,y^2\,-\,2y\:=\:-9$

$\text{Then: }\:x^2\,-\,2x\,+\,1\,+\,y^2\,-\,2y\,+\,1 \:=\:-9\,+\,1\,+\,1$

$\text{Hence: }\:(x\,-\,1)^2\,+\,(y\,-\,1)^2 \:=\:-7$
[color=beige]. . . . . . . . . . . . . . . . . . . . . . . .[/color]$\uparrow$
[color=beige]. . . . . . . . . . . . . . [/color]$Now\text{ we know the radius!}$

$\text{The circle has an imaginary radius and does not exist.}$

$\text{Consider the circle: }\:x^2\,+\,y^2\,-\,6x\,+\,2y \:=\:-6 \;\text{ . . . which exists.}$

$\text{W\!e have: }\:x^2\,-\,6x\,+\,y^2\,+\,2y \:=\:-6$

$\text{Then: }\:x^2\,-\,6x\,+\,9\,+\,y^2\,+\,2y\,+\,1 \:=\:-6\,+\,9\,+\,1$

$\text{Hence: }\:(x\,-\,3)^2\,+\,(y\,+\,1)^2 \:=\:4$

$\text{This circle has center }(3,\,-1)\text{ and radius }2.$

Got it?

 August 12th, 2012, 05:07 PM #3 Newbie   Joined: Aug 2012 Posts: 5 Thanks: 0 Re: The Radius of a Circle Yes, of course! Thank you so much. When I apply this to the problem that I'm actually working on, I see that the circle really does exist. Andrew

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# radius of a circle is negative

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