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

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

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Answer 1 · 2015-12-29T11:47:33

Use the substitutions $x = r cos (θ)$ $x = rcos(theta)$ and $y = r sin (θ)$ $y = rsin(theta)$

Explanation:

Use the substitutions $x = r cos (θ)$ $x = rcos(theta)$ and $y = r sin (θ)$ $y = rsin(theta)$
Then the equation becomes
$r^{2} {cos}^{2} (θ) + r^{2} {sin}^{2} (θ) = 9$ $r^2cos^2(theta) + r^2sin^2(theta) = 9$
$r^{2} ({cos}^{2} (θ) + {sin}^{2} (θ)) = 9$ $r^2(cos^2(theta) + sin^2(theta) ) = 9$
Because we know that ${cos}^{2} (θ) + {sin}^{2} (θ) = 1$ $cos^2(theta) + sin^2(theta) = 1$
this gives $r^{2} = 9$ $r^2 = 9$ and hence $r = 3$ $r= 3$

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

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