Question

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

Answer 1

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

Answer 2

i think

Answer 3

explain it to me just incase

Answer 4

\(\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?

Answer 5

not sure it rings :D

Answer 6

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

Answer 7

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

Answer 8

aaaahhhh i dont know it

Answer 9

\(\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

Answer 10

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

Answer 11

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

Answer 12

so how is this related to the question

Answer 13

well.. lemme group it a bit

Answer 14

okay :)

Answer 15

\(\bf x^2-2x+y^2-6y=26\implies (x^2-2x)+(y^2-6y)=26
\\ \quad \\
(x^2-2x+{\color{red}{ \square }}^2)+(y^2-6y+{\color{red}{ \square }}^2)=26\)

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?

Answer 16

nop :(

Answer 17

well, think about it

recall, the middle term, given here
is 2 * both of the other terms

Answer 18

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

Answer 19

is the factor for the first one -1,2

Answer 20

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

Answer 21

yes but the coordinate which gives you the first trionomial

Answer 22

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

Answer 23

\(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ \square }}^2)=26\)

how about the one for the "y" group?

Answer 24

is it 1 as well?

Answer 25

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

Answer 26

is it 3

Answer 27

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 \(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ 3 }}^2)-{\color{red}{ 1 }}^2-{\color{red}{ 3 }}^2=26\)

Answer 28

okay

Answer 29

now
let us simplify that
keeping in mind that, the groups, are not perfect square trinomials
thus

\(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ 3 }}^2)-{\color{red}{ 1 }}^2-{\color{red}{ 3 }}^2=26
\\ \quad \\
(x^2-2x+1^2)+(y^2-6y+3^2)-1-9=26
\\ \quad \\
(x^2-2x+1^2)+(y^2-6y+3^2)-10=26
\\ \quad \\
(x^2-2x+1^2)+(y^2-6y+3^2)=26+10
\\ \quad \\
(x^2-2x+1^2)+(y^2-6y+3^2)=36
\\ \quad \\
(x-1)^2+(y-3)^2=36\implies (x-1)^2+(y-3)^2=6^2\)

Answer 30

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?

Answer 31

your explanation is perfect but i dont seem to get it

Answer 32

i saw what you did there

Answer 33

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

see them now?

Answer 34

but i dont see the center coordinates

Answer 35

oooooohhhhhhh

Answer 36

okay :)

Answer 37

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

Answer 38

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

Answer 39

yw

Answer 40

i appriciate your time and good explanation

Answer 41

np