How do you find the rectangular equation for $r = - 6 sin θ$ $r=-6sintheta$ ?

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Answer 1 · 2016-11-12T17:09:30

Please see the explanation for steps leading to the equation of a circle:

${(x - 0)}^{2} + {(y - - 3)}^{2} = 3^{2}$ $(x - 0)^2 + (y - -3)^2 = 3^2$

Multiply both sides of the equation by r:

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

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

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

Add $6 y + k^{2}$ $6y + k^2$ to both sides:

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

Use the right side of the pattern ${(y - k)}^{2} = y^{2} - 2 k y + k^{2}$ $(y - k)^2 = y^2 - 2ky + k^2$ , to complete the square for the y terms:

$y^{2} - 2 k y + k^{2} = y^{2} + 6 y + k^{2}$ $y^2 - 2ky + k^2 = y^2 + 6y + k^2$

$- 2 k y = 6 y$ $-2ky = 6y$

$k = - 3$ $k = -3$ and $k^{2} = 9 = 3^{2}$ $k^2 = 9 = 3^2$

Replace the y terms with the left side of the pattern but with $k = - 3$ $k = -3$ :

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

To put this into the standard form for a circle, substute $3^{2}$ $3^2$ for $k^{2}$ $k^2$ (not 9) )and insert a $- 0$ $-0$ in the x term:

${(x - 0)}^{2} + {(y - - 3)}^{2} = 3^{2}$ $(x - 0)^2 + (y - -3)^2 = 3^2$

This is a circle with its center at $(0, - 3)$ $(0, -3)$ and a radius of 3.

How do you find the rectangular equation for r=-6sinthetar=−6sinθr=-6sintheta?