Question

Find the center and radius of the circle whose equation is x^2+2x+y^2+4y-11=0

Answer 1

the general equation of a circle is

\[x^2+y^2+2gx+2fy+c=0\]

Answer 2

where (-g,-f) is the centre

Answer 3

*center

Answer 4

solution is posted on your other post

Answer 5

I got the center as 1,2 and the radius 16

Answer 6

radius=\(\sqrt{g^2+f^2-c}\)

Answer 7

Seems @dumbcow already answered your question :)

Answer 8

maybe easier to write in the form
\[(x-h)^2+(y-k)^2=r^2\] get the center is \((h,k)\) and radius is \(r\)
do you know how to complete the square?

Answer 9

ok thank you

Answer 10

\[x^2+2x+y^2+4y-11=0\]
\[x^2+2x+y^2+4y=11\]
\[(x+1)^2+(y+2)^2=11+1^2+2^2=16\]

Answer 11

i got the center as (1,2) and radius 16

Answer 12

careful here

Answer 13

the form is
\[(x-h)^2+(y-k)^2=r^2\] and you have
\[(x+1)^2+(y+2)^2=16\]

Answer 14

oh! (2,1)

Answer 15

that means the center is \((-1,-2)\) and the radius is \(4\)

Answer 16

no, not (2,1) but rather (-1,-2)

Answer 17

Oh i see my mistake thank you

Answer 18

here is a good reference to practice and learn completing the square

http://www.purplemath.com/modules/sqrcircle.htm