$r = \sin(\theta) + \cos(\theta)$
$r^2 = r\sin(\theta) + r\cos(\theta)$
$x^2 + y^2 = x+ y$
$x^2 + y^2 - x - y = 0$
Is this right?
$r = \sin(\theta) + \cos(\theta)$
$r^2 = r\sin(\theta) + r\cos(\theta)$
$x^2 + y^2 = x+ y$
$x^2 + y^2 - x - y = 0$
Is this right?
Yes it is, your approach is very good, but you can go one step further:
$$x^2-x+y^2-y=0\iff \left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac12$$
is a circle of center $\left(\dfrac12,\dfrac12\right)$ and radius $\dfrac1{\sqrt2}$. It passes through the origin.