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!