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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)
(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
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)
can you please show me the steps PLEASE!!
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?
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...
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?
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
you see the (x-4) part in your problem?
so is that how it came??
thats exactly how it came; ... :)
so if the equation is (x-5)^2+(y-1)^2=25 then r=5 and centre= (5,1)???
you got it, that is absolutely correct..... good job :)
so there is no complicated algebra we gotta do??
so what if the equation is (x-(1/2))^2+(y+(3/4)^2=(1/4)
then just pull out the fraction stuff as your answer, x = 1/2 y = -3/4 radius = sqrt(1/4) = 1/2
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 . . . .