Question

show that the equation x^2+y^2+x+k(x^2+y^2-2x+y-1)=0 represents a circle, and find the center and the radius

Answer 1

what i did so far was.. \[x^2+y^2+x+k(x^2+y^2-2x+y-1)=0\]\[x^2+y^2+x+kx^2+ky^2-2kx+ky-k=0\]\[x^2+kx^2+y^2+ky^2+x-2kx+ky-k=0\]\[(1+k)x^2+(1+k)y^2+(1-2k)x+ky-k=0\]thats wat i did but im stuck there...

Answer 2

@freckles

Answer 3

I think complete the square. Rearrange a little first
\[[(1+k)x^2+(1-2k)x]+[(1+k)y^2+ky]=k\]
divide by (1 + k)
\[(x^2+x)+(y^2+\frac{ k }{ 1+k })=\frac{ k }{ 1+k }\]

Answer 4

\[(x+\frac{ 1 }{ 2 })^2+\left( y+\frac{ k }{ 2(1+k) } \right)^2=\frac{ k }{ 1+k }+\frac{ 1 }{ 4 }+\frac{ k^2 }{ 4(1+k)^2 }\]

Answer 5

that's really nasty, but once you combine the stuff on the right, the square root of it will be the radius

Answer 6

\[(x-h)^2+(y-k)^2=r^2\]
(h, k) = center and r = radius

Answer 7

shdnt it be \[x^2+\frac{ 1-2k }{ 1+k }x+y^2+\frac{ k }{ 1+k }y=\frac{ k }{ 1+k }\]

Answer 8

@peachpi

Answer 9

that's correct so it becomes

\[(x + \frac{1 - 2k}{2(1 + k)})^2 + (y + \frac{k}{2(1 + k)})^2 = \frac{k}{1 + k} + (\frac{1 - 2k}{2(1 + k)})^2 + (\frac{k}{2(1 + k})^2 \]

then just simplify the right hand side

Answer 10

oh yeah, sorry. good catch

Answer 11

thanks guys