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

ok so i did

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

where do i go from here

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- May 25th 2009, 08:24 PMtomatoesconverting polar equations into rectangular form
r =3cos(theta) +sin(theta)

ok so i did

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

where do i go from here - May 25th 2009, 08:40 PMpickslides
$\displaystyle r =3cos(\theta) +sin(\theta)$

multiply each side by r giving

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

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

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

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

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