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

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

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Answer 1 · 2016-04-15T18:22:32

$r^{2} + 8 r cos θ + 16 = 0$ $r^2+8rcostheta+16=0$

Explanation:

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

Use the equations:
$x^{2} + y^{2} = r^{2}, x = r cos θ, y = r sin θ$ $x^2+y^2=r^2, x=rcos theta, y=rsin theta$

$0 = r^{2} + 8 r cos θ + 16$ $0=r^2+8rcostheta+16$

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

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

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