Finding the Center and Radius of a Circle: Analytic Geometry I

AI Thread Summary
The discussion focuses on finding the center and radius of the circle represented by the equation x²+y²-4x+2y+6=0. Participants clarify that the circle is not defined because the radius squared (r²) equals -1, which is impossible. They discuss the method of completing the square to rewrite the equation in the standard form of a circle. One user emphasizes the importance of understanding the process rather than relying solely on memorized formulas to avoid mistakes. The conversation concludes with a mutual acknowledgment of the method's effectiveness.
kLownn
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Homework Statement


Find the centre and radius of the circle x²+y²-4x+2y+6=0

I have the solution. The circle is no defined because r² = -1 is impossible.
But... how do I even DO that equation to get the answer -1?!
 
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You complete the squares so you can write it in the form (x+a)^2+(y+b)^2=c. Do you know how to do that?
 
Dick said:
You complete the squares so you can write it in the form (x+a)^2+(y+b)^2=c. Do you know how to do that?

I learned how to do that while I was in school.. I forget how to do it now.. >_<
 
Take the x part. You've got x^2-4x. If I add something to that it will become a perfect square of the form (x-a)^2. What's 'a'? What do you have to add?
 
Dick said:
Take the x part. You've got x^2-4x. If I add something to that it will become a perfect square of the form (x-a)^2. What's 'a'? What do you have to add?

Oh! Is "a" 2x?
(x-2x)² ?
 
No, no. (x-a)^2=x^2-2ax+a^2, yes? If you match that up with x^2-4x, the 4 must be the 2a, as I see it. Think back to when you did this before.
 
Dick said:
No, no. (x-a)^2=x^2-2ax+a^2, yes? If you match that up with x^2-4x, the 4 must be the 2a, as I see it. Think back to when you did this before.

Ohh, I see now.. so I just do the same thing for y?
 
kLownn said:
Ohh, I see now.. so I just do the same thing for y?

Sure.
 
Dick said:
Sure.

Thank you so much! :)
 
  • #10
kLownn said:

Homework Statement


Find the centre and radius of the circle x²+y²-4x+2y+6=0

I have the solution. The circle is no defined because r² = -1 is impossible.
But... how do I even DO that equation to get the answer -1?!

it can actually be express as x²+y²+2fx+2gy+c=0

whrby C(-f,-g) and radius is \sqrt{g²+f²-c}
 
  • #11
icystrike said:
it can actually be express as x²+y²+2fx+2gy+c=0

whrby C(-f,-g) and radius is \sqrt{g²+f²-c}
No, it isn't. It is much better to actually do the "complete the square" rather than memorize formulas: so you don't make silly mistakes like that.
 
  • #12
HallsofIvy said:
No, it isn't. It is much better to actually do the "complete the square" rather than memorize formulas: so you don't make silly mistakes like that.

yep (: noted.
 
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