Draw the locus of points which satisfy the equation (1) Draw the locus of points $(x,y)$ which satisfy the following equation.
$bx^{3}+y^{3}+x^{2}y+bxy^{2}-4abxy-2ab^{2}x^{2}-2ay^{2}+b\left( a^{2}b^{2}+a^{2}-1\right)x+\left( a^{2}b^{2}+a^{2}-1\right) y=0 $
(2) When the locus can be drawn with a single stroke, show the relation between $a$ and $b$.
Thank you very much :)
 A: This equation can be factorized like this:
$$
\big(b \!\cdot\! x + y\big)\big(x^2 + y^2 + a^2 \!\cdot\! b^2 + a^2  - 1 - 2 \!\cdot\! a \!\cdot\! (b\!\cdot\!x + y)\big) = 0
$$
So either 
$$
b \!\cdot\! x + y = 0 \label{1} \tag{1}
$$  or 
$$ x^2 + y^2 + a^2 \!\cdot\! b^2 + a^2  - 1 - 2 \!\cdot\! a \!\cdot\! (b\!\cdot\!x + y) = 0 \label{2}\tag{2}
$$
We can transform $\eqref{2}$
$$
(x - a\!\cdot\!b)^2 + (y - a)^2 = 1
$$
Therefore, in general case, the locus is the reunion between a line and a circle. 
For this locus to be drawn with a single stroke all we can do is to force the equation $\eqref{1}$ not to give a line as solution, so $b = 0$ and the origin is on the circle, or force the line to intersect the circle. 
So, if $b = 0$ then the only solution for equation $\eqref{1}$ is $x = 0$, $y = 0$ (the origin). As I said, the one stroke condition implies the circle must contain the origin, so $a = 1$ or $a = -1$.
Now, if $b$ != $0$, then replacing $y = -b * x$ in equation $\eqref{2}$ will give us the relation between $a$ and $b$.
