How do you graph $r = - 6 sin θ$ $r=-6sintheta$ ?

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Answer 1 · 2016-10-01T01:29:20

It is a circle with center at (0, -3) and radius, 3:
$x^{2} + {(y - - 3)}^{2} = 3^{2}$ $x^2 + (y - -3)^2 = 3^2$

Multiply both sides by r:

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

Substitute $x^{2} + y^{2}$ $x^2 + y^2$ for r and y for $r sin (θ)$ $rsin(theta)$

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

$x^{2} + y^{2} + 6 y = 0$ $x^2 + y^2 + 6y = 0$

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

$x^{2} + {(y + 3)}^{2} = 3^{2}$ $x^2 + (y + 3)^2 = 3^2$

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

How do you graph r=-6sinthetar=−6sinθr=-6sintheta?