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

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Homework Help Overview

The discussion revolves around finding the center and radius of a circle defined by the equation x²+y²-4x+2y+6=0, which falls under the subject area of analytic geometry. Participants explore the implications of the equation, particularly the assertion that the radius squared equals -1, indicating a potential issue with the circle's definition.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of completing the square to rewrite the equation in a standard form. There are questions about how to identify the necessary values to complete the square for both x and y terms. Some express uncertainty about the process and seek clarification on specific steps.

Discussion Status

The discussion is active, with participants offering guidance on completing the square and questioning assumptions about the equation's implications. There is a mix of attempts to clarify the method and some participants expressing a preference for understanding the process rather than relying on memorized formulas.

Contextual Notes

Some participants note that the equation can be expressed in a different form involving parameters f and g, but there is disagreement on the effectiveness of this approach compared to completing the square. The original poster expresses confusion about how to derive the conclusion that r² = -1.

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 [tex]\sqrt{g²+f²-c}[/tex]
 
  • #11
icystrike said:
it can actually be express as x²+y²+2fx+2gy+c=0

whrby C(-f,-g) and radius is [tex]\sqrt{g²+f²-c}[/tex]
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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