Question

find the center and radius of the cirlce x^2+y^2-6x+4y+9=0

Answer 1

complete the square twice
\[x^2-6x+y^2+4y=-9\] is a start
then for the \(x\) terms you get
\[(x-3)^2+y^2+4y=-9+9=0\]
can you do it for the \(y\) terms?

Answer 2

i actually kind of figured out but for the center im hesistating that if its -3 and 2

Answer 3

you got it backwards, it is \((3,-2)\)

Answer 4

general form is
\[(x-h)^2+(y-k)^2=r^2\]

Answer 5

thank you! do you know anything about canonical equation?