Question

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

Answer 1

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)

Answer 2

(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

Answer 3

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)

Answer 4

can you please show me the steps PLEASE!!

Answer 5

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?

Answer 6

yes

Answer 7

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...

Answer 8

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?

Answer 9

-4

Answer 10

no -3

Answer 11

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

Answer 12

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

Answer 13

so is that how it came??

Answer 14

thats exactly how it came; ... :)

Answer 15

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

Answer 16

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

Answer 17

so there is no complicated algebra we gotta do??

Answer 18

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

Answer 19

then just pull out the fraction stuff as your answer,

x = 1/2 y = -3/4

radius = sqrt(1/4) = 1/2

Answer 20

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 . . . .