Question

College Algebra: reviewing for final exam!
Name the center and radius of the circle:
(equation in comment below)

Answer 1

\[x ^{2}+y ^{2}-4x+6y-3=0\]

Answer 2

Latex is the killer, my connection snapped.

Answer 3

Mine too, I had to refresh twice.

Answer 4

\(\large\color{black}{ x^2+y^2-4x+6y-3=0 }\)
lets rearrange it a bit,
\(\large\color{black}{ x^2-4x+y^2+6y-3=0 }\)
then add 3 to both sides,
\(\large\color{black}{ x^2-4x+y^2+6y=3 }\)

and we are looking at:
\(\large\color{black}{ \color{blue}{x^2-4x}+\color{red}{y^2+6y}=3 }\)

Answer 5

1) what number do you add to the BLUE, to make it a perfect square?
1) what number do you add to the RED, to make it a perfect square?

Answer 6

the second question is number 2, my bad.

Answer 7

4 would be 2 and 6 would be ....

Answer 8

I apologize, I don;t understand.

Answer 9

The perfect square for 4 is 2. But I'm not sure if 6 has a perfect square

Answer 10

no no, we are adding a number that would make a trinomial a perfect square.

Answer 11

Ohhh. Wouldn't you add 4 to 4 to be 16 and add 6 to 6 to be 36?

Answer 12

I mean multiply

Answer 13

oh I love how this site disconnects me.

Answer 14

andf again just now.

Answer 15

It was happening to me a lot this morning. I get disconnected, but when I refresh, the page stays refreshing for about 5 minutes to more.

Answer 16

\(\large\color{black}{ \color{blue}{x^2-2x+1}+\color{red}{y^2+8y+16}=4\color{blue}{+1} \color{red}{+16} }\) I mean this.

Answer 17

Yes, okay, back to the prob. do you see what I am doing above?

Answer 18

Not really. I'm trying to understand how you got those numbers

Answer 19

I mean.
\(\large\color{black}{ \color{blue}{x^2-2x}+\color{red}{y^2+8y}=4 }\)

\(\large\color{black}{ \color{blue}{x^2-2x+1}+\color{red}{y^2+8y+16}=4\color{blue}{+1} \color{red}{+16} }\)

Answer 20

I am adding numbers.
1 to the blue, and 16 to the red.

Answer 21

because "1" would make the blue polynomial a perfect square trinomial, and so the same will do 16 to the red.

Answer 22

Where did the 2 and 8 come from?

Answer 23

you mean the 1 and 16?

Answer 24

2 and 8, are just another example.

Answer 25

do you understand how I did the 2 steps in my last reply in red and blue?

Answer 26

Nope, you lost me

Answer 27

sorry.

Answer 28

Do you know how to complete the square?

Answer 29

Wouldn't that involve \[(-\frac{ b }{ 2 })^{2}\] ?

Answer 30

yes.

Answer 31

this number you just posted is the number needed to complete the square for a trinomial.

Answer 32

Ohhhh, I got it!!

Answer 33

So, when we do this....
\(\large\color{black}{ \color{blue}{x^2-2x}+\color{red}{y^2+8y}=4 }\)

\(\large\color{black}{ \color{blue}{x^2-2x+1}+\color{red}{y^2+8y+16}=4\color{blue}{+1} \color{red}{+16} }\)

you see what I am doing right?

Answer 34

Yes I do.

Answer 35

but it is
\(\large\color{black}{ (\frac{b}{2})^2 }\)
not minus.
where b is coefficient, in
\(\large\color{black}{ x^2+bx+c }\)

Answer 36

Oh, my professor told me it had a negative

Answer 37

it does not. unless the "b" in your trinomial is negative then you say namely,
(-2/2)^2

Answer 38

Of course would still be positive answer lolx

Answer 39

Why does god do this to me?

Answer 40

(._.)

Answer 41

okay, back to your problem.
\(\large\color{black}{ x^2+y^2-4x+6y-3=0 }\)
rearrangement.
\(\large\color{black}{ \color{blue}{x^2-4x}+\color{red}{y^2+6y}-3=0 }\)

\(\large\color{black}{ \color{blue}{x^2-4x}+\color{red}{y^2+6y}=3 }\)

Answer 42

what numbes, do you add to the blue and to the red, to make both trinomials perfect squares/

Answer 43

Alright, I would add 4 to the blue and 9 to the red.

Answer 44

Yup:)

Answer 45

Woohoo! :)

Answer 46

\(\large\color{black}{ \color{blue}{x^2-4x\underline{+4}}+\color{red}{y^2+6y\underline{+9}}=3\color{blue}{\underline{+4}}\color{red}{\underline{+9}} }\)

Answer 47

See, and I am not changing the value of both sides, right?

Answer 48

like
a=b
then
a+3=b+3

type of thing.

Answer 49

Now, you need to write it in your normal form that we so looked for:)

Answer 50

\[(x-4)^{2}+(y-6)^{2}=16\]
I felt like this is what the normal form is from the last question I asked

Answer 51

your Xs is right. check the y.

Answer 52

Oops
\[(x-4)^{2}+(y+6)^{2}=16\]

Answer 53

yes, +6 lol;)

Answer 54

I'll be leaving in about 5 minutes... so have a good night and a good math.

Answer 55

Well, you have a good night as well :)
Thank you so much for your assistance :)

Answer 56

the main thing is that you get the steps of doing it. There are going to be times when completing the square gets very awkward. I even now flip on some problems. but YW, and have agood one!

Answer 57

(o.o) Um, I haven't finished answering the question lolx
@SolomonZelman

Answer 58

I have to go to bad:)

Answer 59

sorry.

Answer 60

Aww, it's okay. I'll see if someone can finish it for me :)
Have a good night, sweet dreams!

Answer 61

@dan815
Do you know how to finish this problem?
As far as I got is \[(x-4)^{2}+(y+6)^{2}=16\]

Answer 62

@dan815
I need to know the center and the radius

Answer 63

you already know it
\[(x-h)^2+(y-k)^2=r^2\]

Answer 64

Compare your equation to this.

Answer 65

(h,k) is the center

Answer 66

Welcome back SolomonZelman!
I did and I know my radius is 4, but I don't know the center. According to my packet, the center is not (4,-6)

Answer 67

you have a Russian name.

Answer 68

and yes, the center is (4,-6).

Answer 69

Thank you (^_^)
Sadly, I'm not Russian, I'm Puerto Rican :D

Answer 70

Why is that sadly? that's nice.

Answer 71

I have a packet my professor gave me and it says the center is (2,-3) and the radius is 4. I already knew the radius because 4*4 is 16

Answer 72

Being part Russian would be cool :)

Answer 73

No it wouldn't be, trust me. Well, in my case it is knowing 3 languages, including English.

Answer 74

I only know 2 :D
I'll ask my professor to see if he had a typo on this problem

Answer 75

why this problem seems good.
and you have a radius and a center.

Answer 76

Yeah, but the answer is SUPPOSE to be (2,-3) and r=4

Answer 77

then there must be something wrong. the answer can't be that.

Answer 78

That's what I'm thinking!

Answer 79

Yes:)
We didn't make any mistakes:)

Answer 80

I'll make sure to ask my professor tomorrow. Thank you so much :)

Answer 81

Anytime;)