I really don't get this lesson question:
Which statement is true for the circle?
http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/0800/0800_G8_Q29a.gif
Its center is at the ordered pair of 1 over 2, minus 1, and the radius is minus five over two units.
Its center is at the ordered pair of minus 1 over 2, 1, and the radius is five over two units.
Its center is at the ordered pair of 1 over 2, minus 1, and the radius is five over two units.
Its center is at the ordered pair of minus 1 over 2, 1, and the radius is minus five over two units.

- anonymous

- katieb

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- anonymous

@amistre64

- anonymous

@jim_thompson5910

- mathstudent55

The link doesn't work

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- anonymous

it works on mine

- anonymous

@AccessDenied

- anonymous

This link works for you because it is on a course website and you have to be logged in to view it, do you know how to take a screenshot?

- anonymous

umm i think lemme try

- anonymous

Okay if you need help, reply back with your computer OS(Operating System, Mac or Windows)

- anonymous

Here's the whole thing as pasted from the lesson

##### 1 Attachment

- phi

a radius is always positive

- anonymous

okay...?

- phi

that rules out 2 of your choices

- anonymous

so it's either B or C, but I still don't know which

- phi

the (x,y) pair at the center will make the (x-a)^2 and (y-b)^2 terms zero
does that make sense?

- anonymous

not really, no...

- phi

usually they say
(x-h)^2 + (y-k)^2 = r^2
and the center is at (h,k) with radius r

- anonymous

how do we get h and k?

- phi

so if you see (x- 1/2) that means the x value of the center is +1/2

- anonymous

okay

- phi

in this case you match your equation to the "generic" one
match
(x-h)^2 + (y-k)^2 = r^2
(x-1/2)^2 + (y + 1)^2 = 25/4

- phi

matching y-k to y+1 you have to rewrite y+1 as y - (-1)
now you see y - k matches y - (-1) and k is -1

- phi

or remember that you want the y value that makes (y+1)^2 zero. that would be y= -1

- anonymous

so the Y is -1?

- anonymous

then it's C? Right?

- phi

yes.

- anonymous

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- phi

thank you

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