How do you convert $x^{2} + {(y - 4)}^{2} = 16$ $x^2+(y-4)^2=16$ to polar form?

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How do you convert $x^{2} + {(y - 4)}^{2} = 16$ $x^2+(y-4)^2=16$ to polar form?

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Answer 1 · 2016-05-03T10:46:56

$r = 8 sin θ$ $r=8sintheta$

Explanation:

If $(r, θ)$ $(r,theta)$ is in polar form and $(x, y)$ $(x,y)$ in Cartesian form the relation between them is as follows:

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ , $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $tan θ = \frac{y}{x}$ $tantheta=y/x$

Hence, $x^{2} + {(y - 4)}^{2} = 16$ $x^2+(y-4)^2=16$ can be written as

$x^{2} + y^{2} - 8 y + 16 = 16$ $x^2+y^2-8y+16=16$ or

$x^{2} + y^{2} - 8 y = 0$ $x^2+y^2-8y=0$ or

$r^{2} - 8 r sin θ = 0$ $r^2-8rsintheta=0$ or

$r (r - 8 sin θ) = 0$ $r(r-8sintheta)=0$ dividing by $r$ $r$

$r - 8 sin θ = 0$ $r-8sintheta=0$ or

$r = 8 sin θ$ $r=8sintheta$

Answer 2 · 2016-05-03T10:54:48

$r = 8 sin θ$ $r=8sintheta$

Explanation:

To convert from Cartesian to Polar coordinates use the following formulae that link them.

$∙ x = r cos θ and y = r sin θ$ $• x=rcostheta" and " y=rsintheta$

$x^{2} + y^{2} - 8 y + 16 = 16 (expanding bracket)$ $x^2+y^2-8y+16=16 " (expanding bracket) "$

$\Rightarrow r^{2} {cos}^{2} θ + r^{2} {sin}^{2} θ - 8 r sin θ + 16 - 16 = 0$ $rArrr^2cos^2theta+r^2sin^2theta-8rsintheta+16-16=0$

then $r^{2} ({cos}^{2} θ + {sin}^{2} θ) - 8 r sin θ = 0$ $r^2(cos^2theta+sin^2theta)-8rsintheta=0$

using the identity: ${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$

$\Rightarrow r^{2} = 8 r sin θ and dividing both sides by r$ $rArr r^2=8rsintheta" and dividing both sides by r "$

$\Rightarrow r = 8 sin θ$ $rArr r=8sintheta$

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How do you convert $x^{2} + {(y - 4)}^{2} = 16$ $x^2+(y-4)^2=16$ to polar form?

2 Answers

Explanation:

Explanation:

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How do you convert $x^{2} + {(y - 4)}^{2} = 16$ $x^2+(y-4)^2=16$ to polar form?