How do you convert $r= -8(cos(x)+(sin(x))$ into cartesian form?

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Answer 1 · 2016-03-23T02:03:28

$(x+4)^2+(y+4)^2=32$

$r=-8(cos x+sinx)$

$r^2=-8r(cosx+sinx)$ -> multiply both sides by r

$r^2=-8rcosx-8rsinx$ ->distribute

$x^2+y^2=-8x-8y$ ->use formula $x=rcos theta, y=rsin theta$

$x^2+8x+y^2+8y=0$

$(x^2+8x+16)+(y^2+8y+16)=16+16$ -> complete the square

$(x+4)^2+(y+4)^2=32$

How do you convert r= -8(cos(x)+(sin(x)) r= -8(cos(x)+(sin(x)) into cartesian form?