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anonymous
 5 years ago
using standard equation to give the radius and center of the circle;
(x4)^2+(y3)^2=16
a little HELP!! plz!!
anonymous
 5 years ago
using standard equation to give the radius and center of the circle; (x4)^2+(y3)^2=16 a little HELP!! plz!!

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that 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
 5 years ago
Best ResponseYou've already chosen the best response.0(x4) 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well you know the standard formula is (xh)^2 + (yk)^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 xcoordinates of the center is 4 and the ycoordinate of the center is 3. So your center would be located at (4,3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you please show me the steps PLEASE!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0say 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0lets 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats 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
 5 years ago
Best ResponseYou've already chosen the best response.0you see the (x4) part in your problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so is that how it came??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats exactly how it came; ... :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if the equation is (x5)^2+(y1)^2=25 then r=5 and centre= (5,1)???

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you got it, that is absolutely correct..... good job :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so there is no complicated algebra we gotta do??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what if the equation is (x(1/2))^2+(y+(3/4)^2=(1/4)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then just pull out the fraction stuff as your answer, x = 1/2 y = 3/4 radius = sqrt(1/4) = 1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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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