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

1. mathstudent55

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

2. anonymous

@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

3. Mertsj

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

4. Mertsj

$(x+1)^2+(y-2)^2=11^2$

5. Mertsj

$x^2+2x+1+y^2-4y+4=121$

6. Mertsj

$x^2+2x+y^2-4y=116$

7. Mertsj

Doesn't seem right, does it?

8. anonymous

no

9. anonymous

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

10. Mertsj

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.

11. Mertsj

So try the third choice.

12. Mertsj

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

13. Mertsj

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

14. Mertsj

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

15. anonymous

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

16. Mertsj

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

17. Mertsj

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

18. anonymous

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

19. Mertsj

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

20. anonymous

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

21. Mertsj

That would be good.

22. anonymous

do you know how I can upload a pic?

23. Mertsj

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

24. anonymous

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

25. Mertsj

I have never done that on this site.

26. anonymous

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

27. Mertsj

ok

28. anonymous
29. anonymous
30. Mertsj

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 .

31. anonymous

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

32. Mertsj

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

33. mathstudent55

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.

34. anonymous

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 :(

35. Mertsj

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

36. anonymous

so b for that one?

37. Mertsj

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

38. anonymous

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

39. Mertsj

yw

40. anonymous

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

41. anonymous

nvm I think i got it