How do you convert $9 = {(x - 7)}^{2} + {(y + 7)}^{2}$ $9=(x-7)^2+(y+7)^2$ into polar form?

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Answer 1 · 2017-07-15T17:18:42

$r^{2} - 14 r (sin θ - cos θ) + 89 = 0$ $r^2-14r(sintheta-costheta)+89=0$

or $r^{2} - 14 \sqrt{2} r sin (θ - \frac{π}{4}) + 89 = 0$ $r^2-14sqrt2rsin(theta-pi/4)+89=0$

The relation between polar coordinates $(r, θ)$ $(r,theta)$ and rectangular coordinates $(x, y)$ $(x,y)$ is given by

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ i.e. $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Hence we can write $9 = {(x - 7)}^{2} + {(y + 7)}^{2}$ $9=(x-7)^2+(y+7)^2$ as

${(r cos θ - 7)}^{2} + {(r sin θ + 7)}^{2} = 9$ $(rcostheta-7)^2+(rsintheta+7)^2=9$

or $r^{2} {cos}^{2} θ - 14 r cos θ + 49 + r^{2} {sin}^{2} θ + 14 r sin θ + 49 = 9$ $r^2cos^2theta-14rcostheta+49+r^2sin^2theta+14rsintheta+49=9$

or $r^{2} - 14 r (sin θ - cos θ) + 89 = 0$ $r^2-14r(sintheta-costheta)+89=0$

or $r^{2} - 14 \sqrt{2} r sin (θ - \frac{π}{4}) + 89 = 0$ $r^2-14sqrt2rsin(theta-pi/4)+89=0$

How do you convert 9=(x-7)^2+(y+7)^29=(x−7)2+(y+7)29=(x-7)^2+(y+7)^2 into polar form?