Polar equation of the form $r = 5\sin(\theta) + 5\cos(\theta)$
The Cartesian equation is of the form $(x-A)^2+(y-B)^2 = R^2$
Find $A,B$, and $R$.
Guess: Let $x = R\cos(\theta) + A$ and $y = R\sin(\theta)+B$. Plug them in and get on the left hand side:
$$50\sin^2(\theta)\cos^2(\theta)+25\cos^4(\theta)+25\sin^4(\theta)+50\sin(\theta)\cos^3(\theta)+50\sin^3(\theta)\cos(\theta)$$
Plug in the right hand side:
$$R^2 = 25 + 50\sin(\theta)\cos(\theta)$$
Then set the 2 sides equal to each other. It looks as if $R = 5$.
Not sure how to go about this further.
Best Answer
I would say
$r = 5\sin \theta + 5\cos \theta,\quad x = r\cos \theta, \quad y = r\sin \theta$
$\Rightarrow r^2 = 5x + 5y\Rightarrow x^2+y^2=5x + 5y\Rightarrow (x-\frac{5}{2})^2+(y-\frac{5}{2})^2=25/2$