Find the center and the radius of the equation below. I don't think my answer is correct.. i got (1,-2) for the center and 1=radius. can someone check for me please? the equation is below

Question

anonymous · Answer

$$4x^2+4y^2-8x+24y+4=0$$

isaiah.feynman · Answer

group the x's and y's together, and complete the square separately.

anonymous · Answer

that's what i did :) but my circle doesn't look.... like a circle. to find more points, don't i just move 1 unite up, down, left, right because of the radius??

anonymous · Answer

Let us first simplify by dividing all terms by 4,. That way you will have x^2 and y^2 with a coefficient of 1

anonymous · Answer

oooh forgot that! thanks @Florcelda !

anonymous · Answer

show me the new expression, and we will go from there, to group x and y terms

anonymous · Answer

new expression is 4(x^2+y^2-2x+6y+1)

anonymous · Answer

=zero

anonymous · Answer

we can divide both sides by 4 giving
x^2 + y^2 - 2x + 6y + 1 = 0

anonymous · Answer

so is my standard form (x-1)^2+(y+3)^2=8??

anonymous · Answer

you can now transpose 1 to the right side and group our x's and y's. Leave a space for the linear terms. We will complete the squares

anonymous · Answer

sorry, i'm moving a little ahead of you :)

anonymous · Answer

= 9

anonymous · Answer

did you found your mistake in getting 8?

anonymous · Answer

oh i see, i forgot to add the +1 from the xterm, thank you! so this makes my circle (1,-3) and radius 3?

anonymous · Answer

Perfect!

anonymous · Answer

thanks! so in order to find other points to make sure.. i just use the radius ( go up, down, left and right 3units from the center?)

anonymous · Answer

Yes

anonymous · Answer

Thank you so much!

anonymous · Answer

My pleasure