Find the parametric curve representation in polar coordinates for $(x-1)^2+(y+1)^2=8$.
Usually equations of form $x^2+y^2=a^2$ have the parametric curve like: $$ c(t)=\langle a\cos t,a\sin t\rangle $$ But I'm struggling to find parametrization for the equation above.
While trying to simplify I got the following: $$ (r\cos t-1)^2+(r\sin t+1)^2=8\\ r^2(\cos^2 t+\sin^2t)+2r\sin t-2r\cos t=6\\ r^2+2r(\sin t-\cos t)=6 $$ But I don't really see how this helps.
On
Think about it geometrically. Your circle is centred on $(p, q)$ so you'll have $x = p + r\cos t$ and $y = q + r\sin t$ which is why what you're attempting isn't working.
Another way of looking at it is to substitute $x'=x-p$ and $y'=y-q$ and parametrizing $(x')^2+(y')^2=a^2$ before substituting back.
You get $x'=a\cos t$ and $y' = a\sin t$ which results in the above.
On
just hint
Taking your last equation, the reduced discriminant is $$\delta=(\cos (t)-\sin (t))^2+6$$ $$=7-\sin (2t) $$
thus
$$r=\cos (t)-\sin (t)\pm \sqrt {7-\sin (2t)}$$
This is the polar equation.
the parametric equations are $$x=1+\sqrt {8}\cos (f) $$ $$y=-1+\sqrt {8}\sin (f) $$ with $0\le f\le 2\pi $.
You can also set $x-1 = \sqrt{8} \cos t$ and $y+1 = \sqrt{8} \sin t$ this will verify the equation and the parametrization becomes $c(t) = \langle 1 + \sqrt{8} \cos t, -1 + \sqrt{8} \sin t\rangle$