Question

College Algebra: I'm reviewing for my Final coming up and I need tons of help. Some of the math problems I'm going through are not making sense. The center of the circle with the following equation is: (comment below)

Answer 1

\[(x-4)^{2}+(y+9)^{2}=25\]

Answer 2

you will likely need logarithms, rules for shifts of functions, circles and other like staff....

Answer 3

Have you like made a big outline of things you need to know?

Answer 4

you just need to know about circles?

Answer 5

\(\large\color{black}{ (x-h)^2+(y-k)^2=r^2 }\)
where \(\large\color{black}{ (h,k) }\) is the center of the circle.

Answer 6

Well, I had a standard form of a circle in mind, but then I started thinking it wouldn't fit. It's
\[(x _{2}-h)^{2}+(y _{2}-k)^{2}=r ^{2}\]

Answer 7

Yeah, there you go lolx

Answer 8

it is not x sub of something, just \(\large\color{black}{ (x-h)^2+(y-k)^2=r^2 }\)

Answer 9

Do you have any problems to work with, or want me to think of an example?

Answer 10

I gave you the equation in the first comment.

Answer 11

Sorry, I was blind.. okay, and what do you want, to extract the center of it?

Answer 12

I believe so, the math problem only said "The center of the circle with the following equation is:"

Answer 13

\(\large\color{black}{ (x\color{blue}{-h})^2+(y\color{red}{-k})^2=r^2 }\)

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

`center is (h,k)`

can you compare the equations?

Answer 14

first, can you tell me the h ?

Answer 15

Basically -4 and 9 is the center of the circle.

Answer 16

no it is the other way,

Answer 17

So, 4 and -9?

Answer 18

yes,


\(\large\color{black}{ (x-\color{blue}{4})^2+(y+9)^2 =25 }\)

\(\large\color{black}{ (x-\color{blue}{4})^2+(y-\color{red}{-9})^2 =25 }\)

\(\large\color{black}{ (x-\color{blue}{4})^2+(x-\color{red}{h})^2 =r^2 }\)

Answer 19

I am re-writing the y+9, as y-(-9), to give it a form of (y-k).
in this case, k is -9.
See?

Answer 20

Yes I do :)

Answer 21

GOod;)

If I were to ask you,
What is the center of the following circle: \(\large\color{black}{ (x+3)^2+(y+4)^2 =16 }\)
what would you tell me?

Answer 22

That the center of the circle is (-3, -4)

Answer 23

yes.

Answer 24

Yes, I got it!! :D

Answer 25

And a center of the above?
\(\large\color{black}{ x^2+(y-2)^2 =16 }\)

Answer 26

Oh, that's tricky.
isn't it (-2, 256) or (1, -2)?

Answer 27

Hint:
\(\large\color{black}{ (x-0)^2+(y-2)^2 =16 }\)

Answer 28

Oh, (0,2) lolx

Answer 29

yes.

Answer 30

And lastly,
\(\large\color{black}{ x^2+y^2 =0 }\) ?

Answer 31

I am tricking you...

Answer 32

(0,0) :)

Answer 33

So i noticed :D

Answer 34

You're also teaching me, which is good. I do better with repetition :)

Answer 35

No.

Answer 36

When
\(\large\color{black}{ x^2+y^2 =0 }\)
comparing to
\(\large\color{black}{ (x-h)^2+(y-k)^2 =r^2 }\)

our radius, the r, is zero. and this is therefore not a circle.

Answer 37

So, no solution right?

Answer 38

yes, it is not a circle.

Answer 39

That was a tricky one

Answer 40

okay, and how about this,
\(\large\color{black}{ (x-2)^2+(y-2)^2 =-2 }\) ?

Answer 41

I'm thinking it's (2,2), but that -2 is making me second guess myself.

Answer 42

a radius can;t be an imaginary number, because if r^2=-2,
then \(\large\color{black}{ r=2i }\).

Answer 43

Ohhh. So it's not a circle because of the radius

Answer 44

it is not a circle again.
The radius has to be greater than zero.

Answer 45

yes, i's not a circle because of the radius.
And lastly,
\(\large\color{black}{ (x+1)^2+y^2=9 }\)

Answer 46

I keep saying and lastly and conttinuing... lol

Answer 47

(-1, 0)
Yeah I just realized that :D

Answer 48

yes.

Answer 49

:)

Answer 50

So, you just need the circles?

Answer 51

Yes, I answered my question :D

Answer 52

You helped me answer and understand it. Thank you very much, I appreciate it :)

Answer 53

\(\large\color{black}{ \smile }\) you welcome!