How do you convert $4 = {(- x - 3)}^{2} + {(2 y - 4)}^{2}$ $4=(-x-3)^2+(2y-4)^2$ into polar form?

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How do you convert $4 = {(- x - 3)}^{2} + {(2 y - 4)}^{2}$ $4=(-x-3)^2+(2y-4)^2$ into polar form?

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Answer 1 · 2016-10-26T20:12:57

Expand the squares:

$4 = x^{2} + 6 x + 9 + 4 y^{2} - 16 y + 16$ $4 = x^2 + 6x + 9 + 4y^2 - 16y + 16$

Combine the constants and group the square terms together:

$x^{2} + 4 y^{2} + 6 x - 16 y + 21 = 0$ $x^2 + 4y^2 + 6x - 16y + 21 = 0$

Substitute $r cos (θ)$ $rcos(theta)$ for x and $r sin (θ)$ $rsin(theta)$ for y:

$({cos}^{2} (θ) + 4 {sin}^{2} (θ)) r^{2} + (6 cos (θ) - 16 sin (θ)) r + 16 = 0$ $(cos^2(theta) + 4sin^2(theta))r^2 + (6cos(theta) - 16sin(theta))r + 16 = 0$

This is a quadratic in r, use the quadratic formula:

$r = \frac{(16 sin (θ) - 6 cos (θ)) \pm \sqrt{{((6 cos (θ) - 16 sin (θ)))}^{2} - 64 ({cos}^{2} (θ) + 4 {sin}^{2} (θ))}}{2 ({cos}^{2} (θ) + 4 {sin}^{2} (θ))}$ $r = {(16sin(theta) - 6cos(theta)) +-sqrt(((6cos(theta) - 16sin(theta)))^2 - 64(cos^2(theta) + 4sin^2(theta)))}/(2(cos^2(theta) + 4sin^2(theta)))$

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How do you convert $4 = {(- x - 3)}^{2} + {(2 y - 4)}^{2}$ $4=(-x-3)^2+(2y-4)^2$ into polar form?

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How do you convert $4 = {(- x - 3)}^{2} + {(2 y - 4)}^{2}$ $4=(-x-3)^2+(2y-4)^2$ into polar form?