Question

Help me understand?

Answer 1

aah the mystery of completing the square

Answer 2

yeah, I'm rusty I guess

Answer 3

\[y^2-4y\] complete the square
half of 4 is 2, so
\[(y-2)^2\] does it
but \\[(y-2)^2=y^2-4y+4\] so you have to add \(4\) to the right

Answer 4

that is why
\[x^2+y^2-4y=0\] is equivalent to
\[x^2+(y-2)^2=4\]

Answer 5

what about the x^2?

Answer 6

that is already complete

Answer 7

it is the square of x right?

Answer 8

maybe it is not clear what the goal is here
you have
\[x^2+y^2-4y=0\] a perfectly good equation
the point is to figure out what this looks like
it is a circle, with center \((0,2)\) and radius 4, but you may not recognize that in the form that it is in

Answer 9

scratch that, the radius is 2

Answer 10

Oh okay, so I'm trying to get it into the standard form of a circle

Answer 11

right

Answer 12

So I only dealt with the y's in the equation because it was "incompelte"?

Answer 13

*incomplete

Answer 14

general form for a circle is
\[(x-h)^2+(y-k)^2=r^2\] and you have '
\[x^2+(y-2)^2=2^2\]

Answer 15

visualize as
\[(x-0)^2+(y-2)^2=r^2\]

Answer 16

you don't mess with the \(x^2\) part because it is just \(x^2\) not say \(x^2-10x\) or \(x^2+6x\)

Answer 17

okay that makes sense. and the number I added--the four--is that added because of the 4y?

Answer 18

\[(y-2)^2=y^2-4y+4\] so when you change \(y^2-4y\) in to \((y-2)^2\) you are adding 4 to the left
that is okay, so long as you add 4 to the right as well

Answer 19

okay thank you!!

Answer 20

yw