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

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

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Answer 1 · 2015-12-06T14:45:29

Substitute $x = r cos θ$ $x = r cos theta$ and $y = r sin θ$ $y = r sin theta$ into the equation to get:

${(r cos θ + 1)}^{2} + {(r sin θ + 1)}^{2} = 4$ $(r cos theta + 1)^2 + (r sin theta + 1)^2 = 4$

Hence:

$r^{2} + 2 (cos θ + sin θ) r - 2 = 0$ $r^2 + 2(cos theta + sin theta)r - 2 = 0$

Explanation:

Substitute $x = r cos θ$ $x = r cos theta$ and $y = r sin θ$ $y = r sin theta$ into the equation to get:

${(r cos θ + 1)}^{2} + {(r sin θ + 1)}^{2} = 4$ $(r cos theta + 1)^2 + (r sin theta + 1)^2 = 4$

This can be reformulated as follows:

$4 = {(r cos θ + 1)}^{2} + {(r sin θ + 1)}^{2}$ $4 = (r cos theta + 1)^2 + (r sin theta + 1)^2$

$= r^{2} {cos}^{2} θ + 2 r cos θ + 1 + r^{2} {sin}^{2} θ + 2 r sin θ + 1$ $= r^2 cos^2 theta + 2r cos theta + 1 + r^2 sin^2 theta + 2r sin theta + 1$

$= r^{2} ({cos}^{2} θ + {sin}^{2} θ) + 2 r (cos θ + sin θ) + 2$ $= r^2 (cos^2 theta + sin^2 theta) + 2r (cos theta + sin theta) + 2$

$= r^{2} + 2 r (cos θ + sin θ) + 2$ $= r^2 + 2r (cos theta + sin theta) + 2$

Subtract $4$ $4$ from both ends to get:

$r^{2} + 2 (cos θ + sin θ) r - 2 = 0$ $r^2 + 2(cos theta + sin theta)r - 2 = 0$

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

1 Answer

Explanation:

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Math

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

1 Answer

Explanation:

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