How do you convert $-2y+1=(x+4)^2+(y-1)^2$ into polar form?

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Answer 1 · 2016-04-15T18:22:32

$r^2+8rcostheta+16=0$

$-2y+1=x^2+8x+16+y^2-2y+1$
$0=x^2+8x+16+y^2-2y+1+2y-1$
$0=x^2+8x+16+y^2$
$0=(x^2+y^2)+8x+16$

Use the equations:
$x^2+y^2=r^2, x=rcos theta, y=rsin theta$

$0=r^2+8rcostheta+16$

How do you convert -2y+1=(x+4)^2+(y-1)^2-2y+1=(x+4)^2+(y-1)^2 into polar form?