How do you convert $1 = {(- x + 4)}^{2} + {(2 y + 9)}^{2}$ $1=(-x+4)^2+(2y+9)^2$ into polar form?

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How do you convert $1 = {(- x + 4)}^{2} + {(2 y + 9)}^{2}$ $1=(-x+4)^2+(2y+9)^2$ into polar form?

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Answer 1 · 2018-03-01T15:40:16

$r^{2} ({cos}^{2} θ + 4 {sin}^{2} θ) - r (4 cos θ - 18 sin θ) + 40 = 0$ $r^2(cos^2theta+4sin^2theta)-r(4costheta-18sintheta)+40=0$

Explanation:

Given:
$1 = {(- x + 4)}^{2} + {(2 y + 9)}^{2}$ $1=(-x+4)^2+(2y+9)^2$

$1 = {(- x + 4)}^{2} + {(2 (y + \frac{9}{2}))}^{2}$ $1=(-x+4)^2+(2(y+9/2))^2$
$1 = {(- x + 4)}^{2} + 4 {(y + \frac{9}{2})}^{2}$ $1=(-x+4)^2+4(y+9/2)^2$

$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

Substituting
$1 = {(- r cos θ + 4)}^{2} + {(2 r sin θ + 9)}^{2}$ $1=(-rcostheta+4)^2+(2rsintheta+9)^2$

$1 = r^{2} {cos}^{2} θ - 8 r cos θ + 16 + 4 r^{2} {sin}^{2} θ + 36 r sin θ + 81$ $1=r^2cos^2theta-8rcostheta+16+4r^2sin^2theta+36rsintheta+81$

$r^{2} ({cos}^{2} θ + 4 {sin}^{2} θ) + r (- 8 cos θ + 36 sin θ) + 81 - 1 = 0$ $r^2(cos^2theta+4sin^2theta)+r(-8costheta+36sintheta)+81-1=0$

$2 r^{2} ({cos}^{2} θ + 4 {sin}^{2} θ) - 2 r (4 cos θ - 18 sin θ) + 2 (40) = 0$ $2r^2(cos^2theta+4sin^2theta)-2r(4costheta-18sintheta)+2(40)=0$

$r^{2} ({cos}^{2} θ + 4 {sin}^{2} θ) - r (4 cos θ - 18 sin θ) + 40 = 0$ $r^2(cos^2theta+4sin^2theta)-r(4costheta-18sintheta)+40=0$

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How do you convert $1 = {(- x + 4)}^{2} + {(2 y + 9)}^{2}$ $1=(-x+4)^2+(2y+9)^2$ into polar form?

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Explanation:

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How do you convert $1 = {(- x + 4)}^{2} + {(2 y + 9)}^{2}$ $1=(-x+4)^2+(2y+9)^2$ into polar form?