Question

find the center and radius of the circle x^2+2x-4y+6=0
will FAN and MEDAL

Answer 1

This is not a circle. Are you sure you copied the equation correctly? You need a y^2 term.

Answer 2

@mathstudent55 That is literally whats on my test. The answer options are:
center (-1,2) radius :11
center (-1,2) radius: sq root of 11
center (1,-2) radius: 11
center (1,-2) radius sq root of 11

Answer 3

Well then let's work backwards from the answers and see which answer works.

Answer 4

\[(x+1)^2+(y-2)^2=11^2\]

Answer 5

\[x^2+2x+1+y^2-4y+4=121\]

Answer 6

\[x^2+2x+y^2-4y=116\]

Answer 7

Doesn't seem right, does it?

Answer 8

no

Answer 9

I am confused
I kind of get what you did but I don't know how to continue it

Answer 10

The next choice has the same center but a different radius. But the radius should be sqrt116. So it doesn't seem to work either.

Answer 11

So try the third choice.

Answer 12

No. It couldn't be (x-1)^2 because that would be -2x

Answer 13

So of the choices given it has to be the first or the second one. And we know it isn't the first one.

Answer 14

Let's look at choice two more closely. I bet it will work.

Answer 15

ok! after could you help me with another one? it's the same but the equation is x^2+8x-6y+9=0
options are
center (4,-3) radius sq root of 34
center (-4,3) radius sq root of 34
center (4,-3) radius 34
center (-4,3) radius 34

Answer 16

\[(x-1)^2+(y+2)^2=(\sqrt{11})^2\]
\[x^2-2x+1+y^2+4y+4=11\]
\[x^2-2x+y^2+4y+5=11\]
\[x^2-2x+y^2+4y=6\]

Answer 17

The question is obviously flawed if you have posted it correctly but since it is multiple choice, b is the best shot you have.

Answer 18

thankyou! I've had so much trouble with that question!

Answer 19

Something is very wrong. Are you sure you posted the question EXACTLY as stated?

Answer 20

yes! I can post a screenshot if you'd like. I just don't know how to upload a picture

Answer 21

That would be good.

Answer 22

do you know how I can upload a pic?

Answer 23

All circles have x^2 and y^2 in their equations.

Answer 24

yeah that's what I thought but this question is really weird and theres a lot similiar to these on my test

Answer 25

I have never done that on this site.

Answer 26

I've seen other people do it...hmmm let me upload the pic to a site and then share the link

Answer 27

ok

Answer 28

http://s1252.photobucket.com/user/Beautiful_Rainbows/media/Screen%20Shot%202015-07-20%20at%207.46.30%20PM_zpssnamuhhc.png.html

Answer 29

[URL= http://s1252.photobucket.com/user/Beautiful_Rainbows/media/Screen%20Shot%202015-07-20%20at%207.46.30%20PM_zpssnamuhhc.png.html ][IMG] http://i1252.photobucket.com/albums/hh562/Beautiful_Rainbows/Screen%20Shot%202015-07-20%20at%207.46.30%20PM_zpssnamuhhc.png [/IMG][/URL]

try this link

Answer 30

It hasn't finished loading yet. I should warn you that my internet often goes down at 11 for some strange reason the Frontier can't seem to figure out. But the way the problem is posted, it has to be either choice b or .

Answer 31

Oh aw alright! What time is it rn where ur at? I chose answer b but I haven't submitted it yet

Answer 32

I see what you mean. For the second problem, let's try choice b. 10:52

Answer 33

The equation of a circle must have both the x-term and the y-term squared.
The equation you were given is not a circle.
I think they left out the y squared term.

Answer 34

aw man thats soon :( oh well...... wait for the second problem? so not the first one right? @mathstudent55 there are many questions like this on my test :(

Answer 35

\[(x+4)^2+(y-3)^2=(\sqrt{34})^2\]
\[x^2+8x+16+y^2-6y+9=34\]
\[x^2+8x+y^2-6y=9\]

Answer 36

so b for that one?

Answer 37

I would choose b and then I would point out to my teacher that these equations cannot be circles.

Answer 38

Oh okay! Thankyou so much for your time and help!

Answer 39

yw

Answer 40

do you know how to award a medal ? I'm not really sure how to haha

Answer 41

nvm I think i got it