Graph x^2+y^2-2y+1=1

[Ironhawk]

I am doing this problem where I have to Graph x^2+y^2-2y+1=1 a circle. After going thru my text book I see your suppose to write the equation in standard form by completing the square. In the example I see in the book it shows the problem having 2 (x's) and 2 (y's) is there a different way for this problem? Or am I on the wrong track.

According to the book I have to put the y's together and the X's together, add to both sides to complete the square.

 

[wjm11]

I am doing this problem where I have to Graph x^2+y^2-2y+1=1 a circle. After going thru my text book I see your suppose to write the equation in standard form by completing the square. In the example I see in the book it shows the problem having 2 (x's) and 2 (y's) is there a different way for this problem? Or am I on the wrong track.

Hello, Ironhawk,

When we “complete the square” we are doing so in order to create a “perfect square trinomial.” You already have one in this problem: y^2 – 2y + 1. This can be factored as (y – 1)^2, so our equation can be written as

x^2 + (y – 1)^2 = 1

The equation for a circle may be written as (x – h)^2 + (y – k)^2 = 1, where (h,k) is the center of the circle. In this case, h = 0, so our circle center is located at (0,1). Make sense?

 

[stapel]

...is there a different way for this problem?

Not that I'm aware of. Completing the square is generally the way to go.

Fortunately, while completing the square is a somewhat annoying process, using it to find the center and radius of a circle is pretty straightforward, once you've practiced a few times. Have fun!

Eliz.

 

[mmm4444bot]

... The equation for a circle may be written as (x – h)^2 + (y – k)^2 = 1 , where (h,k) is the center of the circle ...

Hello Iron Hawk:

I believe that wjm intended to write r^2 instead of 1, where r is the radius of the circle.

Cheers,

~ Mark

 

[mmm4444bot]

... while completing the square is a somewhat annoying process ...

GASP!

(Somebody just dissed one of my favorite manual chores.)

:wink: