Question

put x^2+y^2-2x+4y+1=0 into the standard form for circles. Show work.

Answer 1

(x-1)^2 + (y + 2)^2 = 2^2

Answer 2

how?

Answer 3

add and subtract 4 and 1

Answer 4

\[x^2-2x+y^2-4y-1=0\]\[(x-1)^2-1+(y+2)^2-4+1=0\]
\[(x-1)^2+(y+2)^2=4\]

Answer 5

Like lana

Answer 6

(x-1)^2 - 1^2 + (y+2)^2 - 2^2 + 1 = 0
(x-1)^2 + (y+2)^2 -4 = 0
C(1,-2) R = 2

Answer 7

i dont get how you guys got these answers

Answer 8

Hm, using the formula
(x-h)^2 + (y+k)^2 = r^2

Answer 9

how do you find h and k tho?

Answer 10

i completed the square
first for x^-2x
then for y^2+4y

Answer 11

what did you get when you compleated the squares?
and how?

Answer 12

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

Answer 13

do you subtract on the opposite side, if not, then whats the point of subtracting?

Answer 14

Use the formula
x^2 + bx = (x+b/2)^2 - (b/2)^2
x^2 - bx = (x-b/2)^2 - (b/2)^2

Answer 15

this is how i complete the square
\[x^2+bx= (x+\frac{b}{2})^2-(\frac{b}{2})^2\]

Answer 16

so rather than subtracting (b/2)^2, can i just add it to the other side?

Answer 17

yeah u can

Answer 18

so suppose we're using this question (which im still confused on), it becomes
(x+1)^2+(y+2)^2=4

Answer 19

after that do i expand the parenthetical's?

Answer 20

Expand!?
(x+1)^2+(y+2)^2=4 thats already the standard form of the circle

Answer 21

that become the standard form then? so center would be (-1,-2)?

Answer 22

Yeah for that circle the centre is (-1,-2) and the radius is 2

Answer 23

ohh, thanks a whole bunch! i get it now!... i think...

Answer 24

lols "i think" tell me if you don't get it i can explain further

Answer 25

im gonna try another similar question on my hw and then check back in here

Answer 26

so, can you tell me how to solve for

Answer 27

center is (1,-2) not (-1,-2)

Answer 28

(x+1)^2+ [(2x+1)+2]^2=4?

Answer 29

Um , for (x+1)^2+(y+2)^2=4 the C(-1,-2) Xd

Answer 30

xD*

Answer 31

wait, now how did you get (1,-2) instead of (-1,-2)?

Answer 32

No, she was looking at your previous question xD

Answer 33

oh lol im sleepy

Answer 34

the one up top?

Answer 35

yeh sory ,, didn know its a new question nw

Answer 36

(x+1)^2+ [(2x+1)+2]^2=4?
Is that right ?

Answer 37

wow, no, i actually need the answer to the original question

Answer 38

ohlol then am not sleepy

Answer 39

I already gave you the answer to the orginal one xD

Answer 40

but yes, what im trying to do in my hw problem is disprove line y=2x+1 intersecting with circle X^2+y^2-2x+4y+1=0

Answer 41

so im a little confused again, did i set the second part up right?

Answer 42

disprove!? what do you mean by that

Answer 43

show that (said line) does not intersect (said equation)

Answer 44

so did i set it up right?

Answer 45

Set up?

Answer 46

so im trying to disprove line y=2x+1 intersecting with circle X^2+y^2-2x+4y+1=0
we got down to the standard form of (x+1)^2+(y+2)^2=4 and y=2x+1, when i plug it in is it supposed to look like this: (x+1)^2+[(2x+1)+2]^2=4?

Answer 47

yeh thats right

Answer 48

ok, i think i got the rest then... hopfully.

Answer 49

wait,

Answer 50

anyways if u need any help am here

Answer 51

what i ment to type was, how am i supposed to expand [(2x+1)+2]^2?

Answer 52

sry, didnt post

Answer 53

\[(2x+3)(2x+3)= 4x^2+12x+9\]

Answer 54

ok, so i add the 2 before the i square?

Answer 55

yeh coz its added

Answer 56

ok, thanks alot

Answer 57

ur welcome=)