Question 1157773
a)

 {{{x^2 + y^2 = 16}}}

to convert to Polar apply: {{{x=r*cos (theta )}}},{{{y=r*sin(theta )}}}

{{{(r*cos (theta ))^2 + (r*sin(theta ))^2 = 16}}}

{{{r^2*cos^2 (theta ) + r^2*sin^2(theta ) = 16}}}

{{{r^2(cos^2 (theta ) + sin^2(theta )) = 16}}}

{{{r^2(1) = 16}}}

So, in your case, the equation becomes simply 

{{{r^2 = 16}}}

This means that the equation represents all the points with distance{{{ 4}}} from the origin, which is a circumference with radius {{{4}}}, centered in the origin.


b) 

{{{x^2 + y^2 = 10}}}

{{{x=r*cos (theta )}}},{{{y=r*sin(theta )}}}

{{{(r*cos (theta ))^2 + (r*sin(theta ))^2 = 10}}}

{{{r^2(cos^2 (theta ) + sin^2(theta )) = 10}}}

{{{r^2(1) = 10}}}

the equation becomes simply 

{{{r^2 = 10}}}


c) 

{{{y = 3}}}

{{{y=r*sin(theta )}}}

{{{r*sin(theta )=3}}}



d) 

{{{y = -4x}}}

{{{r*sin(theta )=-4r*cos (theta )}}}

{{{r*sin(theta )+4r*cos (theta )=0}}}

{{{r(sin(theta) + 4cos(theta)) = 0}}}


e) 

{{{3xy = 5}}}

{{{3r*cos (theta )*r*sin(theta )=5}}}

{{{3r^2*cos (theta )*sin(theta )=5}}}