How do you convert #r = 5 cos (t)# into cartesian form?

1 Answer
Bill K.
Jan 4, 2016

Multiply both sides by #r# and substitute #r^{2}=x^{2}+y^{2}# and #rcos(t)=x# to get #x^{2}+y^{2}=5x#. Using the method of Completing the Square , this can be transformed to #(x-5/2)^{2}+y^{2}=25/4=(5/2)^{2}# to see that the curve is a circle of radius #5/2# centered at the point #(x,y)=(5/2,0)#.

Explanation:

Cartesian form means using rectangular coordinates rather than polar coordinates. Most typically, the "conversion equations" would be written with a #theta# rather than at #t# as:

#r^{2}=x^{2}+y^{2},\ x=rcos(theta),\ y=rsin(theta)#.

After the substitution mentioned above, the equation #x^{2}+y^[2}=5x# can be rewritten as #x^{2}-5x+y^{2}=0#, or #x^2-5x+(-5/2)^2+y^{2}=(-5/2)^2#, or

#(x-5/2)^{2}+y^{2}=(5/2)^{2}#

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