anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jdoe0001
  • jdoe0001
do you know what a "perfect square trinomial" is?
anonymous
  • anonymous
i think
anonymous
  • anonymous
explain it to me just incase

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jdoe0001
  • 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
  • anonymous
not sure it rings :D
anonymous
  • anonymous
well its the one where you identify A B and C right?
jdoe0001
  • 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
  • anonymous
aaaahhhh i dont know it
jdoe0001
  • 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
  • anonymous
so you just combined them but it still means the same thing right
jdoe0001
  • 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
  • anonymous
so how is this related to the question
jdoe0001
  • jdoe0001
well.. lemme group it a bit
anonymous
  • anonymous
okay :)
jdoe0001
  • 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
  • anonymous
nop :(
jdoe0001
  • jdoe0001
well, think about it recall, the middle term, given here is 2 * both of the other terms
jdoe0001
  • 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
  • anonymous
is the factor for the first one -1,2
jdoe0001
  • jdoe0001
well, the last number in a perfect trinomial is positive, so... can't be -1 for one
anonymous
  • anonymous
yes but the coordinate which gives you the first trionomial
jdoe0001
  • 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
  • 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
  • anonymous
is it 1 as well?
jdoe0001
  • 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
  • anonymous
is it 3
jdoe0001
  • 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
  • anonymous
okay
jdoe0001
  • 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
  • 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
  • anonymous
your explanation is perfect but i dont seem to get it
anonymous
  • anonymous
i saw what you did there
jdoe0001
  • jdoe0001
\(\bf (x-{\color{brown}{ 1}})^2+(y-{\color{blue}{ 3}})^2={\color{purple}{ 6}}^2\) see them now?
anonymous
  • anonymous
but i dont see the center coordinates
anonymous
  • anonymous
oooooohhhhhhh
anonymous
  • anonymous
okay :)
jdoe0001
  • jdoe0001
so the center of the circle is at (1,3) and the radius is 6
anonymous
  • anonymous
hhhhmmmm thank you so much.you really broke it down for me
jdoe0001
  • jdoe0001
yw
anonymous
  • anonymous
i appriciate your time and good explanation
jdoe0001
  • jdoe0001
np

Looking for something else?

Not the answer you are looking for? Search for more explanations.