Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit.
x2 −2x + y2 − 6y = 26

- anonymous

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- schrodinger

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- jdoe0001

do you know what a "perfect square trinomial" is?

- anonymous

i think

- anonymous

explain it to me just incase

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## More answers

- jdoe0001

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

- anonymous

not sure it rings :D

- anonymous

well its the one where you identify A B and C right?

- jdoe0001

well https://s3.amazonaws.com/ck12bg.ck12.org/curriculum/108348/thumb_540_50.jpg <--- that show it more or less
is a trinomial, so it has 3 terms
the terms on the extremes, left and right
are squared
the term in the middle is
the multiplication of 2 times the other two terms, without the exponent

- anonymous

aaaahhhh i dont know it

- jdoe0001

\(\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow
\end{array}\qquad
% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array}\)
notice, "a" and "b" are squared, on the extreme sides
the middle is 2 * both, without the exponent

- anonymous

so you just combined them but it still means the same thing right

- jdoe0001

yes, you'd combine them to a binomial raised at 2
and yes, if you expand the binomial, you'd get back the trinomial

- anonymous

so how is this related to the question

- jdoe0001

well.. lemme group it a bit

- anonymous

okay :)

- jdoe0001

\(\bf x^2-2x+y^2-6y=26\implies (x^2-2x)+(y^2-6y)=26
\\ \quad \\
(x^2-2x+{\color{red}{ \square }}^2)+(y^2-6y+{\color{red}{ \square }}^2)=26\)
so... we grouped it first
and then, we'd want to get a "perfect square trinomial" for those groups
any ideas what the missing fellow is there in each?

- anonymous

nop :(

- jdoe0001

well, think about it
recall, the middle term, given here
is 2 * both of the other terms

- jdoe0001

so the missing number, is really hidden there, in the middle term
now if you just factor out the middle term, the missing one will show up

- anonymous

is the factor for the first one -1,2

- jdoe0001

well, the last number in a perfect trinomial is positive, so... can't be -1 for one

- anonymous

yes but the coordinate which gives you the first trionomial

- jdoe0001

but let us use say +1 instead
the middel term is 2x
2 * x * 1 = 2x
so the other terms are "x" and "1"
and 2 * both is "2x"
so, yes, one is 1

- jdoe0001

\(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ \square }}^2)=26\)
how about the one for the "y" group?

- anonymous

is it 1 as well?

- jdoe0001

well, let's see there
the middle term in the "y" group is 6y
2 * y * 1 \(\ne 6y\)
so, no dice, 2 * y is, 2y, but 2y * 1, is not 6y
so is not 1

- anonymous

is it 3

- jdoe0001

2 * y * 3 = 6y <-- yeap, is 3
now keep in mind that
all we're doing is borrowing from our very good fellow Mr Zero, 0
so if we ADD \(1^2\ and \ 3^2\) we also have to SUBTRACT them as well
so we end up with \(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ 3 }}^2)-{\color{red}{ 1 }}^2-{\color{red}{ 3 }}^2=26\)

- anonymous

okay

- jdoe0001

now
let us simplify that
keeping in mind that, the groups, are not perfect square trinomials
thus
\(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ 3 }}^2)-{\color{red}{ 1 }}^2-{\color{red}{ 3 }}^2=26
\\ \quad \\
(x^2-2x+1^2)+(y^2-6y+3^2)-1-9=26
\\ \quad \\
(x^2-2x+1^2)+(y^2-6y+3^2)-10=26
\\ \quad \\
(x^2-2x+1^2)+(y^2-6y+3^2)=26+10
\\ \quad \\
(x^2-2x+1^2)+(y^2-6y+3^2)=36
\\ \quad \\
(x-1)^2+(y-3)^2=36\implies (x-1)^2+(y-3)^2=6^2\)

- jdoe0001

a circle equation looks like \(\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2
\qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad
radius={\color{purple}{ r}}\)
now, can you see the center coordinates and the radius in yours?

- anonymous

your explanation is perfect but i dont seem to get it

- anonymous

i saw what you did there

- jdoe0001

\(\bf (x-{\color{brown}{ 1}})^2+(y-{\color{blue}{ 3}})^2={\color{purple}{ 6}}^2\)
see them now?

- anonymous

but i dont see the center coordinates

- anonymous

oooooohhhhhhh

- anonymous

okay :)

- jdoe0001

so the center of the circle is at (1,3) and the radius is 6

- anonymous

hhhhmmmm thank you so much.you really broke it down for me

- jdoe0001

yw

- anonymous

i appriciate your time and good explanation

- jdoe0001

np

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