How do you convert $r = 3 (4 cos (t))$ $r = 3(4cos(t))$ into rectangular form?

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Answer 1 · 2016-11-12T10:25:47

$x^{2} + y^{2} = 12 x$ $x^2+y^2=12x$

The relation between polar coordinates $(r, t)$ $(r,t)$ and Cartesian coordinates $(x, y)$ $(x,y)$ is given by

$x = r cos t$ $x=rcost$ , $y = r sin t$ $y=rsint$ and $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

Hence, $r = 3 (4 cos t) \Leftrightarrow r^{2} = 12 r cos t$ $r=3(4cost)hArrr^2=12rcost$ or $x^{2} + y^{2} = 12 x$ $x^2+y^2=12x$

i.e. $(x^{2} - 12 x + 36) + {(y - 0)}^{2} = 6^{2}$ $(x^2-12x+36)+(y-0)^2=6^2$

or ${(x - 6)}^{2} + {(y - 0)}^{2} = 6^{2}$ $(x-6)^2+(y-0)^2=6^2$

which is nothing but a circle with center at $(6, 0)$ $(6,0)$ and radius $6$ $6$ , whose graph is as follows:
graph{x^2+y^2=12x [-16, 16, -8, 8]}

How do you convert r = 3(4cos(t))r=3(4cos(t))r = 3(4cos(t)) into rectangular form?