How do you convert $r = 2 sin θ + cos θ$ $r=2sin theta + cos theta$ into rectangular form?

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Answer 1 · 2017-02-25T16:29:14

The equation is ${(x - \frac{1}{2})}^{2} + {(y - 1)}^{2} = \frac{5}{4}$ $(x-1/2)^2+(y-1)^2=5/4$

To convert from polar coordinates $(r, θ)$ $(r, theta)$ to rectangular coordinates $(x, y)$ $(x,y)$ , we use the following equations

$x = r cos θ$ $x=rcostheta$ , $\Rightarrow$ $=>$ , $cos θ = \frac{x}{r}$ $costheta=x/r$

$y = r sin θ$ $y=rsintheta$ , $\Rightarrow$ $=>$ , $sin θ = \frac{y}{r}$ $sintheta=y/r$

$x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Therefore,

$r = 2 sin θ + cos θ$ $r=2sintheta+costheta$

$r = 2 \cdot \frac{y}{r} + \frac{x}{r}$ $r=2*y/r+x/r$

$r^{2} = 2 y + x$ $r^2=2y+x$

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

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

$x^{2} - x + \frac{1}{4} + y^{2} - 2 y + 1 = 1 + \frac{1}{4}$ $x^2-x+1/4+y^2-2y+1=1+1/4$

${(x - \frac{1}{2})}^{2} + {(y - 1)}^{2} = \frac{5}{4}$ $(x-1/2)^2+(y-1)^2=5/4$

This is the equation of a circle, center $(\frac{1}{2}, 1)$ $(1/2,1)$ and radius $= \frac{\sqrt{5}}{2}$ $=sqrt5/2$

How do you convert r=2sin theta + cos thetar=2sinθ+cosθr=2sin theta + cos theta into rectangular form?