How do you convert $x^{2} + {(y - 3)}^{2} = 9$ $x^2+(y-3)^2=9$ to polar form?

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Answer 1 · 2017-04-17T05:15:28

Please see the explanation.

Given: $x^{2} + {(y - 3)}^{2} = 9$ $x^2+(y-3)^2=9$

expand the square:

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

Substitute $r^{2} for x^{2} + y^{2}$ $r^2" for "x^2+y^2$ :

$r^{2} - 6 y = 0$ $r^2-6y=0$

Substitute $r sin (θ)$ $rsin(theta)$ for y:

$r^{2} - 6 r sin (θ) = 0$ $r^2-6rsin(theta)=0$

We can discard a factor of r because the root $r = 0$ $r = 0$ is duplicated at $θ = 0$ $theta= 0$

$r - 6 sin (θ) = 0$ $r-6sin(theta)=0$

$r = 6 sin (θ)$ $r = 6sin(theta)$

How do you convert x2+(y−3)2=9x^2+(y-3)^2=9 to polar form?