Lagrange not returning all critical points? I'm trying to identify the critical points of $f(x,y)=x^3+xy^2-4xy$ on the constraint $(x+1)^2+(y-2)^2\leq1$.
Setting up $\nabla f = \lambda\nabla g$ we have
x: $3x^2+y^2-4y=\lambda(2x+2) \implies \lambda=\tfrac{3x^2+y^2-4y}{2x+2}$
y: $2xy-4x=\lambda(2y-4) \implies \lambda=\tfrac{2xy-4x}{2y-4}=\tfrac{x(2y-4)}{2y-4}=x$
Setting the lambdas equal to each other we get
$\tfrac{3x^2+y^2-4y}{2x+2}=x \implies 3x^2+y^2-4y=2x^2+2x \implies -x^2+2x=y^2-4y$
Now completing the square gives
$-((x^2-2x+1)-1)=(y^2-4y+4)-4 \implies -(x-1)^2+1=(y-2)^2-4$
$\implies (x-1)^2+(y-2)^2=5$
Creating a system of equations with our original constraint gives
$(x+1)^2+(y-2)^2=1$
$(x-1)^2+(y-2)^2=5$
$\implies 1-(x+1)^2=5-(x-1)^2 \implies x=-1$
And plugging back into the constraint gives
$((-1)+1)^2+(y-2)^2=1 \implies y=1,3$
So Lagrange has given the points $(-1,1)$ and $(-1,3)$. However, it's pretty obvious geometrically that critical points exist at $(0,2)$ and $(-2,2)$ (one can also verify that they satisfy $\nabla f=\lambda\nabla g$), and it seems like you can find these points with one of the circles $(x+1)^2+(y-2\pm2)^2=5$, given how it intersects the constraint. But to get a circle like this you would have to alter the inputs to the Lagrange, something I'm pretty sure you can't do, so I'm stuck. Is there some property of Lagranges that allows you to alter their inputs?

To make our lives easier we can rewrite $f$ as
$$f(x,y) = x(x^2+y^2-4y) = x(x^2+(y-2)^2-4)$$
which gives us the system of equations
$$\begin{cases}3x^2+(y-2)^2-4 = 2\lambda(x+1) \\ x(y-2) = \lambda(y-2) \\ (x+1)^2+(y-2)^2=1\end{cases}$$
Now if you assume $y\neq 2$ then $x=\lambda$, giving the points which are the intersections of the circles
$$\begin{cases}(x-1)^2+(y-2)^2=5 \\ (x+1)^2+(y-2)^2 = 1\end{cases} \implies (x,y) = (-1,1),(-1,3)$$
But if $y=2$ that would leave us with the points
$$(x+1)^2 = 1 \implies (x,y) = (0,2),(-2,2)$$
which we have to consider separately because otherwise we are just dividing by zero. And we must consider the extrema of the function as well
$$\begin{cases}3x^2+(y-2)^2 = 4 \\ x(y-2) = 0 \\ \end{cases} \implies (x,y) = (0,0),(0,4),\left(\pm\frac{2}{\sqrt{3}},0\right)$$
but none of these are in the region  of consideration.
