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

1 Answer
Bill K.
Jan 4, 2016

Multiply both sides by rr and substitute r^{2}=x^{2}+y^{2}r2=x2+y2 and rcos(t)=xrcos(t)=x to get x^{2}+y^{2}=5xx2+y2=5x. Using the method of Completing the Square , this can be transformed to (x-5/2)^{2}+y^{2}=25/4=(5/2)^{2}(x−52)2+y2=254=(52)2 to see that the curve is a circle of radius 5/252 centered at the point (x,y)=(5/2,0)(x,y)=(52,0).

Explanation:

Cartesian form means using rectangular coordinates rather than polar coordinates. Most typically, the "conversion equations" would be written with a thetaθ rather than at tt 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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