x^2+y^2-2x+2y=23 in the form of (x-a)^2+(y-b)^2=c

Dealing with the equation of a circle!!!

2. Hi renii, the word 'quick' in a forum title doesn't alert the helpers as much as you think it would.

I can help you with your question, but first do you know how to complete the square?

3. yup i do kno how to complete the sqaure

4. Complete the square on this equation, lump all your constants together and move them to the RHS

5. Great, you will be able to finish this problem in a heart beat then.

Show me your skills with completing the square with these two problems,

$\displaystyle \displaystyle x^2-2x$

and

$\displaystyle \displaystyle y^2+2y$

6. I guess that wasn't quick enough!

(I have this image of a guy sitting in an exam, desparately texting!)

7. Did anybody actually answer this for you? First, rewrite as (x^2 - 2x) +(y^2 + 2y) = 23 To complete the square we need a constant term in each of the sets of parentheses that is equal to one half of the linear term, squared. In both cases this is the number 1 So you then have (x^2 - 2x +1) + (y^2 +2y +1) = 25 (addint 2 to each side. Both the terms in parentheses are now perfect squares, so your are left with (x - 1)^2 + (y + 1)^2 = 25 This is the equation of a circle of radius 5, with center at (1,-1). Graph it and see!

8. Originally Posted by professorcool
Did anybody actually answer this for you? First, rewrite as (x^2 - 2x) +(y^2 + 2y) = 23 To complete the square we need a constant term in each of the sets of parentheses that is equal to one half of the linear term, squared. In both cases this is the number 1 So you then have (x^2 - 2x +1) + (y^2 +2y +1) = 25 (addint 2 to each side. Both the terms in parentheses are now perfect squares, so your are left with (x - 1)^2 + (y + 1)^2 = 25 This is the equation of a circle of radius 5, with center at (1,-1). Graph it and see!
It is expected that students complete their own work. The helpers here are NOT SUPPOSED to answer the question for the OP. They provided the hints necessary for the OP to be able to complete the question him/herself.

9. Originally Posted by renii
yup, i do know how to complete the square
Problem solved!

Alternatively...

$\displaystyle (x-a)^2+(y-b)^2=x^2-2ax+a^2+y^2-2by+b^2$

$\displaystyle \Rightarrow\ a=1,\;\;\;b=-1$