ANSWER
[tex]{x}^{2} + {y}^{2} = 16[/tex]
EXPLANATION
We want to eliminate the parameter from the equation:
[tex]x = 4 \cos(t)...(1)[/tex]
and
[tex]y = 4 \sin(t)...(2)[/tex]
Square both sides of the first equation to get,
[tex] {x}^{2} = (4 \cos(t) )^{2} [/tex]
[tex]{x}^{2} = 16 \cos^{2} (t)...(3)[/tex]
Squaring the second equation gives;
[tex] {y}^{2} = {(4 \sin(t) )}^{2} [/tex]
[tex]{y}^{2} = 16 \sin^{2} (t)...(4)[/tex]
Add equation (3) and (4).
This implies that;
[tex] {x}^{2} + {y}^{2} = 16 \cos^{2} (t) + 16 \sin^{2} (t)[/tex]
Factor the right hand side
[tex] {x}^{2} + {y}^{2} = 16( \cos^{2} (t) + \sin^{2} (t))[/tex]
Recall that;
[tex] \cos^{2} (t) + \sin^{2} (t) = 1[/tex]
[tex] {x}^{2} + {y}^{2} = 16( 1)[/tex]
[tex] {x}^{2} + {y}^{2} = 16[/tex]
