Question

For the equation of a circle x^2+y^2+6y=27, complete the square to put it in the form(x-h)^2+(y-k)^2=r^2, then give center and radius.

Answer 1

@zepdrix

Answer 2

@EmmaMink

Answer 3

@jim_thompson5910

Answer 4

have you covered "perfect square trinomials" yet?

or know what a perfect square trinomial is

Answer 5

i forgot it.

Answer 6

hmm

Answer 7

It's been a while since I've done this

Answer 8

well, you may want to brush up on them before doing the exercise then
check your book for the section on "perfect square trinomial" or sometimes also called "perfect square"

Answer 9

would look like \(\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array}\)

Answer 10

or \(\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow
\end{array}\qquad

% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array}\)

for the positive or negative one

Answer 11

this might help
http://www.regentsprep.org/regents/math/algtrig/ate12/completesq.htm

Answer 12

ok