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- anonymous

using standard equation to give the radius and center of the circle;
(x-4)^2+(y-3)^2=16
a little HELP!! plz!!

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- anonymous

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- amistre64

that 16 tells us the radius is sqrt(16) = 4
the center of the circle is shown by whatever those numbers are that are messing around with your x and y parts.
center is at(4,3)

- amistre64

(x-4) says that it was originally at another location but we moved it to get a normal; reading.... we moved it by -4 to get it back to 0, so it was originally at 0+4 = 4

- anonymous

Well you know the standard formula is (x-h)^2 + (y-k)^2 = r^2.
In this case, r^2=16 so r=4.
H and K are the x- and y- coordinates for the circle respectively, so you know from your equation that the x-coordinates of the center is 4 and the y-coordinate of the center is 3. So your center would be located at (4,3)

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- anonymous

can you please show me the steps PLEASE!!

- amistre64

say you wanna look at something that is on a shelf, what do you do? you pick it up, move it over to where you can get a good lok at it, and when your done you put it back right?

- anonymous

yes

- amistre64

well, when we look at circles, or any equation, we want to observe them from (x=0,y=0) so if they are over someplace else we gotta move them...

- amistre64

lets say this circle is centered at (4,0) how far do we move it to get it to (0,0) so that we can look at it?

- anonymous

-4

- anonymous

no -3

- amistre64

thats correct, so we make a note to ourselves going, we moved x -4, so that when we want to put it back....we can remember where we got it from.
-4 was correct

- amistre64

you see the (x-4) part in your problem?

- anonymous

so is that how it came??

- amistre64

thats exactly how it came; ... :)

- anonymous

so if the equation is
(x-5)^2+(y-1)^2=25
then r=5
and centre= (5,1)???

- amistre64

you got it, that is absolutely correct..... good job :)

- anonymous

so there is no complicated algebra we gotta do??

- anonymous

so what if the equation is
(x-(1/2))^2+(y+(3/4)^2=(1/4)

- amistre64

then just pull out the fraction stuff as your answer,
x = 1/2 y = -3/4
radius = sqrt(1/4) = 1/2

- anonymous

ok so now i have a graph and i have to give the coordinates of the centre, the radius, and the equation of the circle
the graph is like this::
. . . 5 . . . .
. . . 4 . . . .
. . . 3 . . . .
. . . 2 . . . .
. . . 1 . . . .
. . . 0 . . . .
. . -1 . . . .
. . -2 . . . .

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