anonymous
  • anonymous
using standard equation to give the radius and center of the circle; (x-4)^2+(y-3)^2=16 a little HELP!! plz!!
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • 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
  • 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
  • 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
  • anonymous
can you please show me the steps PLEASE!!
amistre64
  • 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
  • anonymous
yes
amistre64
  • 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
  • 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
  • anonymous
-4
anonymous
  • anonymous
no -3
amistre64
  • 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
  • amistre64
you see the (x-4) part in your problem?
anonymous
  • anonymous
so is that how it came??
amistre64
  • amistre64
thats exactly how it came; ... :)
anonymous
  • anonymous
so if the equation is (x-5)^2+(y-1)^2=25 then r=5 and centre= (5,1)???
amistre64
  • amistre64
you got it, that is absolutely correct..... good job :)
anonymous
  • anonymous
so there is no complicated algebra we gotta do??
anonymous
  • anonymous
so what if the equation is (x-(1/2))^2+(y+(3/4)^2=(1/4)
amistre64
  • amistre64
then just pull out the fraction stuff as your answer, x = 1/2 y = -3/4 radius = sqrt(1/4) = 1/2
anonymous
  • 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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