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anonymous
 one year ago
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
 one year ago
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know what a "perfect square trinomial" is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0explain it to me just incase

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \begin{array}{cccccllllll} {\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow \end{array}\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ a}}^2&  &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}}  {\color{blue}{ b}})^2&\leftarrow \end{array}\) does it ring a bell?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not sure it rings :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well its the one where you identify A B and C right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well 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
 one year ago
Best ResponseYou've already chosen the best response.0aaaahhhh i dont know it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\begin{array}{cccccllllll} {\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow \end{array}\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ a}}^2&  &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}}  {\color{blue}{ b}})^2&\leftarrow \end{array}\) notice, "a" and "b" are squared, on the extreme sides the middle is 2 * both, without the exponent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you just combined them but it still means the same thing right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, you'd combine them to a binomial raised at 2 and yes, if you expand the binomial, you'd get back the trinomial

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how is this related to the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well.. lemme group it a bit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf x^22x+y^26y=26\implies (x^22x)+(y^26y)=26 \\ \quad \\ (x^22x+{\color{red}{ \square }}^2)+(y^26y+{\color{red}{ \square }}^2)=26\) 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
 one year ago
Best ResponseYou've already chosen the best response.0well, think about it recall, the middle term, given here is 2 * both of the other terms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0is the factor for the first one 1,2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, the last number in a perfect trinomial is positive, so... can't be 1 for one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but the coordinate which gives you the first trionomial

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf (x^22x+{\color{red}{ 1 }}^2)+(y^26y+{\color{red}{ \square }}^2)=26\) how about the one for the "y" group?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, 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
 one year ago
Best ResponseYou've already chosen the best response.02 * 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 \(\bf (x^22x+{\color{red}{ 1 }}^2)+(y^26y+{\color{red}{ 3 }}^2){\color{red}{ 1 }}^2{\color{red}{ 3 }}^2=26\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now let us simplify that keeping in mind that, the groups, are not perfect square trinomials thus \(\bf (x^22x+{\color{red}{ 1 }}^2)+(y^26y+{\color{red}{ 3 }}^2){\color{red}{ 1 }}^2{\color{red}{ 3 }}^2=26 \\ \quad \\ (x^22x+1^2)+(y^26y+3^2)19=26 \\ \quad \\ (x^22x+1^2)+(y^26y+3^2)10=26 \\ \quad \\ (x^22x+1^2)+(y^26y+3^2)=26+10 \\ \quad \\ (x^22x+1^2)+(y^26y+3^2)=36 \\ \quad \\ (x1)^2+(y3)^2=36\implies (x1)^2+(y3)^2=6^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a 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
 one year ago
Best ResponseYou've already chosen the best response.0your explanation is perfect but i dont seem to get it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i saw what you did there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf (x{\color{brown}{ 1}})^2+(y{\color{blue}{ 3}})^2={\color{purple}{ 6}}^2\) see them now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but i dont see the center coordinates

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the center of the circle is at (1,3) and the radius is 6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hhhhmmmm thank you so much.you really broke it down for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i appriciate your time and good explanation
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