Question

Find the radius and center of the circle.
equation: x^2+y^2-8y-4=-16.

& there is no just x term. I don't know why..

Answer 1

that means h is 0...

Answer 2

like center is at (0,1) or something..

Answer 3

first you need to convert it into standard form
standard form with center (a,b) and radius r is
\[\Large (x-a)^2+(y-b)^2=r^2\]
x^2+y^2-8y-4=-16
add and subtract 16 on left side
\[\Large x^2+y^2-8y+16-16-4=-16\]
simplify
\[\Large (x)^2+(y-4)^2=4\]
can you tell me now the center and radius by comparing with
\[\Large (x-a)^2+(y-b)^2=r^2\]

Answer 4

Why do you add & subtract 16 on the left side? Isn't that redundant?

Answer 5

he likes to confuse himself ;D

Answer 6

to make complete square of (y-4)^2
if you expand it you will get the original equation back.

Answer 7

ok if you are not comfortable with that.
can i write
x^2+y^2-8y-4=-16.
as
x^2+y^2-8y+16=4

??

Answer 8

The center is (0,4) and the radius is 2. x^2+(y-4)^2=4

Answer 9

@xThatOneKidx do not provide just answer.
you need this.
http://openstudy.com/code-of-conduct

Answer 10

@sami-21 Okay, well I kind of understand. A little hazy, though.

Answer 11

Lulz my bad