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

Question

1878 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-23T02:03:28

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

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

$r^{2} = - 8 r (cos x + sin x)$ $r^2=-8r(cosx+sinx)$ -> multiply both sides by r

$r^{2} = - 8 r cos x - 8 r sin x$ $r^2=-8rcosx-8rsinx$ ->distribute

$x^{2} + y^{2} = - 8 x - 8 y$ $x^2+y^2=-8x-8y$ ->use formula $x = r cos θ, y = r sin θ$ $x=rcos theta, y=rsin theta$

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

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

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

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