## 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

1. jdoe0001

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

2. anonymous

i think

3. anonymous

explain it to me just incase

4. 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?

5. anonymous

not sure it rings :D

6. anonymous

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

7. 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

8. anonymous

aaaahhhh i dont know it

9. 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

10. anonymous

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

11. 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

12. anonymous

so how is this related to the question

13. jdoe0001

well.. lemme group it a bit

14. anonymous

okay :)

15. 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?

16. anonymous

nop :(

17. jdoe0001

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

18. 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

19. anonymous

is the factor for the first one -1,2

20. jdoe0001

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

21. anonymous

yes but the coordinate which gives you the first trionomial

22. 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

23. 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?

24. anonymous

is it 1 as well?

25. 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

26. anonymous

is it 3

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

28. anonymous

okay

29. jdoe0001

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

30. 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?

31. anonymous

your explanation is perfect but i dont seem to get it

32. anonymous

i saw what you did there

33. jdoe0001

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

34. anonymous

but i dont see the center coordinates

35. anonymous

oooooohhhhhhh

36. anonymous

okay :)

37. jdoe0001

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

38. anonymous

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

39. jdoe0001

yw

40. anonymous

i appriciate your time and good explanation

41. jdoe0001

np