Question

Find the center and radius of the circle.

3x^2+3y^2-12x-24+12=0
Could u please explain in steps

Answer 1

\( \huge 3x^2+3y^2-12x-\underbrace{24}_{24y}+12=0 \)
is this what you meant.. ? :)

Answer 2

@tylerjh still here?

Answer 3

first of all, you need to know its formula.
\( \LARGE (x-h)^2+(y-k)^2=r^2\)
where Center \(\large h,k\;\;\;\)and radius \(r\)

Answer 4

\[\LARGE 3x^2+3y^2-12x-24y+12=0\]
\[\LARGE (3x^2-12x)+(3y^2-24y)=-12\]
\[\LARGE 3(x^2-4x)+3(y^2-8y)=-12\]
\[\LARGE \frac{3(x^2-4x)}{3}+\frac{3(y^2-8y)}{3}=-\frac{12}{3 }\]
\[\LARGE \frac{\cancel{3}(x^2-4x)}{\cancel3}+\frac{\cancel 3(y^2-8y)}{\cancel 3}=-4\]
\[\LARGE (x^2-4x)+(y^2-8y)=-4\]
what do we need to do here to complete the square.. ? :)

Answer 5

@tylerjh I'm not going to do your homework for you... if you don't collaborate, I'll leave !

Answer 6

im very sorry i left the computer to help my mother with moving. Sorry man. I really am. That wont happen again

Answer 7

Ohh never mind.
I thought you were sitting there watching me do it.
Ok
To complete the square you have to do this:

so you need to add 4 and -4 to x's part (so the value won't change.. 4-4=0 )
so we have:
\[\LARGE (x^2-4x+4)-4+(y^2-8y)=-4\]
now we'll do the same with y's
\(\LARGE \frac{-8}{2}=-4\longrightarrow (-4)^2=16\)
so we'll add 16 and -16
\[\Large (x^2-4x+4)-4+(y^2-8y+16)-16=-4\]
now we move -16 and -4 to the right side...
\[\Large (x^2-4x+4)+(y^2-8y+16)=-4+16+4 \]
and we sum\simplify them...
\[\Large (x^2-4x+4)+(y^2-8y+16)=\cancel{-4}+16\cancel{+4} \]