How do you convert $4 = {(x - 4)}^{2} + {(y - 5)}^{2}$ $4=(x-4)^2+(y-5)^2$ into polar form?

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How do you convert $4 = {(x - 4)}^{2} + {(y - 5)}^{2}$ $4=(x-4)^2+(y-5)^2$ into polar form?

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Answer 1 · 2016-07-21T08:05:11

$r^{2} - 8 r cos θ - 10 r sin θ + 37 = 0$ $r^2-8rcostheta-10rsintheta+37=0$

Explanation:

Relation between polar coordinates $(r, θ)$ $(r,theta)$ and rectangular coordinates is given by $x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ , $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $tan θ = \frac{y}{x}$ $tantheta=y/x$ .

Hence $4 = {(x - 4)}^{2} + {(y - 5)}^{2}$ $4=(x-4)^2+(y-5)^2$ can be written as

$x^{2} - 8 x + 16 + y^{2} - 10 y + 25 = 4$ $x^2-8x+16+y^2-10y+25=4$ or

$x^{2} + y^{2} - 8 x - 10 y + 37 = 0$ $x^2+y^2-8x-10y+37=0$ or

$r^{2} - 8 r cos θ - 10 r sin θ + 37 = 0$ $r^2-8rcostheta-10rsintheta+37=0$

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How do you convert $4 = {(x - 4)}^{2} + {(y - 5)}^{2}$ $4=(x-4)^2+(y-5)^2$ into polar form?

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Explanation:

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How do you convert $4 = {(x - 4)}^{2} + {(y - 5)}^{2}$ $4=(x-4)^2+(y-5)^2$ into polar form?