Question

what are the center of the radius of the circle described by the equation (x=1)^2+(y-7)^2=36

Answer 1

(x-a)^2+(y-b)^2=r^2
so
r=root(36)

Answer 2

centre =(a,b)
so centre= (1,7)

Answer 3

@hdross

Answer 4

the anwser was 6 not 36

Answer 5

what are the coordinate of the center of the circle described by the equation x^2+(y+5)^2=16

Answer 6

as he said, the r = root(36) \[r = \sqrt{36} = 6\]because \[6*6=36\]

Formula for a circle is \[(x-h)^2+(y-k)^2 = r^2\]with radius \(r\) and center \((h,k)\)

You have \[x^2 + (y+5)^2 = 16\]You can make it look the same by rewriting it as\[(x+0)^2+(y+5)^2 = 16\]Do you agree that squaring \[(x+0)^2 = x^2\]If not, watch:
\[(x+0)^2 = (x+0)(x+0) = x(x+0) + 0(x+0) \]\[(x+0)^2 =x*x + x*0 + 0*x + 0*0\]\[ (x+0^2= x^2\]

Answer 7

nuts, last line should be \[(x+0)^2 = x^2\]

Answer 8

To line them up:
\[(x-h)^2+(y-k)^2 = r^2\]\[(x+0)^2 + (y+5)^2 = (\sqrt{16})^2\]You can see that for them to be equal, you must have \[-h = +0\]\[-k=+5\]\[r = \sqrt{16}\]or
\[h = 0\]\[k=-5\]\[r=4\]giving you a center of \((0,-5)\) and a radius of \(4\).

Any questions about what I did?