converting polar equations into rectangular form

**tomatoes** · May 25th 2009, 08:24 PM

r =3cos(theta) +sin(theta)

ok so i did

r^2= (3cos(theta) +sin(theta))^2

where do i go from here

**pickslides** · May 25th 2009, 08:40 PM

Originally Posted by tomatoes

r =3cos(theta) +sin(theta)

ok so i did

r^2= (3cos(theta) +sin(theta))^2

where do i go from here

$r =3cos(\theta) +sin(\theta)$

multiply each side by r giving

$r^2 =r(3cos(\theta) +sin(\theta))$

now as $r^2 = x^2+y^2$ the equation becomes

$x^2+y^2 =r\times 3cos(\theta) +r\times sin(\theta)$

$x^2+y^2 = 3x +y$

$x^2+y^2 - 3x - y = 0$

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