Convert this equation to polar form and solve for r

Question: (((x-4)^2)/16) - ((y^2)/16) = 1

What I tried:
(x^2 - 8x + 16) - y^2 = 16
(rcost)^2 - 8x + 16 - (rsint)^2 = 16
then the 16's cancel. I don't know what to do from here? help please!

continue
r²cos²t-8rcost-r²sin²t=0
factoring out the r²
r²(cos²t-sin²t)=8rcost
since cos²t-sin²t= cos 2t
replace it
r²(cos2t)=8rcost
divide both sides by r
r(cos2t)=8cost
r=8cost/cos2t
now you have r one the left HS, and the T on the RHS