$\psi = x^2+(y-2)^2+\lambda(x^2-y^2-1)$
Establish Lagrange Equations
$$\psi_x = 2x+\lambda(2x)=0 $$
$$\psi_y = 2y-4+\lambda(-2y)=0$$
$$\psi_\lambda = x^2-y^2-1=0$$
$\psi_x*y$
$\psi_y*x$
$\psi_x=2xy+2\lambda xy=0$
$\psi_y=2xy-4x-2\lambda xy=0$
Adding and Simplifying
$4xy-4x=0$
$xy-x=0$
$y=1$
Therefore plugging back into $\psi_\lambda$,
$x^2-1-1=0$
$x=\pm\sqrt2$
Stationary points are: $(\sqrt2,1),(-\sqrt2,1)$
Is this the correct method, and also, how would you find the maximum and minimum? Am I correct in saying that the constraint here is a unit circle and thus the maximum is 1 and the minimum is 1?
For Lagrange multipliers, $$\nabla f(x,y)=\lambda\nabla g(x,y)$$ meaning that $$f_x(x,y)\mathbf{i}+f_y(x,y)\mathbf{j}=\lambda g_x(x,y)\mathbf{i}+\lambda g_y(x,y)\mathbf{j}$$ This gives us two equations: $$f_x(x,y)=\lambda g_x(x,y),\quad f_y(x,y)=\lambda g_y(x,y)$$ We also know from the equation for $g(x,y)$ that $$y=\pm\sqrt{x^2-1}$$
In this case, $$f_x(x,y)=2x,\quad f_y(x,y)=2y-4$$ and $$g_x(x,y)=2x,\quad g_y(x,y)=-2y$$ Therefore, $$2x=\lambda 2x,\quad 2y-4=-\lambda2y$$ From the first equation, we have $\lambda=1$.
From there, it is easy to find $y$ and $x$. However, I don't get the same results that you did.
I've never yet seen Lagrange multipliers used the way you used them - writing some function $\psi$ as $$\psi=f(x)+\lambda (g(x)-1)$$ I'm actually not sure why you included the $-1$ in there. If you go straight to $$f(x,y)=\lambda\nabla g(x,y)$$ then it shouldn't influence the problem until you connect $x$ and $y$. Bringing it in earlier will cause some problems for you.