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

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Answer 1 · 2018-05-01T00:27:35

The Cartesian form of the equation is $(x-4)^2+(y+4)^2=32$ .

I assume that the problem was supposed to have $theta$ 's instead of $x$ 's.

To convert the equation into Cartesian form, first, expand the multiplication, and then multiply everything by $r$ :

$r=8(costheta-sintheta)$

$r=8costheta-8sintheta$

$r^2=8rcostheta-8rsintheta$

Now, use the substitutions $rcostheta=x$ , $rsintheta=y$ , and $r^2=x^2+y^2$ :

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

Now complete the square:

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

$x^2-8x+16+y^2+8y+16=0+16+16$

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

The equation is a circle with center $(4,-4)$ and radius $4sqrt2$ (which is about $5.66$ ). Here's what the graph looks like:

graph{(x-4)^2+(y+4)^2=32 [-12.3, 20.74, -11.75, 4.27]}

Hope this helped!

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