Eliminate the parameter t. find a rectangular equation for the plane curve defined by the parametric equations.
x = 6 cos t, y = 6 sin t; 0 ≤ t ≤ 2π
a. x^2 - y^2 = 6; -6 ≤ x ≤ 6
b. x^2 - y^2 = 36; -6 ≤ x ≤ 6
c. x^2 + y^2 = 36; -6 ≤ x ≤ 6

For the answer to this question, I'll show you the two ways,

Take square 
x^2 = 36cos^2t 
y^2 = 36sin^2t , add both 

x^2+y^2= 36( cos^2t+sin^2t) 
x^2+y^2= 36 (1) 
x^2+y^2=36 a circle with radius R^2=36 , ie, R=+-6 , so -6<x<6 ANSWER C 

The other way , 
cos t = x/6 in a right triangle , adjacent side is x , hypotenuse is 6 
so opposite side is b^2= 6^2-x^2 
b^2 = 36-x^2 
thus, sin t = b/6 , 
sint = sqrt ( 36-x^2) /6 
We know that 
y = 6sint 

y = 6 sqrt (36-x^2) /6 
y = sqrt (36-x^2) 
taking square 
x^2+y^2=36