$$\min x^2+y^2 $$ $$\text{s.t.}\ \ (x-2)^2+(y-3)^2\le 4 \ \ \ \text{and} \ \ \ x^2=4y$$
Please explicitly solve this question by using Lagrange multiplier method.
I accept $(x-2)^2+(y-3)^2=4$
And $$L(X,y,\lambda)=x^2+y^2+\lambda_1 [(x-2)^2+(y-3)^2-4]+\lambda_2[x^2-4y]$$
But I cannot calculate Optimal x y and $\lambda,\mu$ . I got ridiculus results:(
One of your conditions is $(x-2)^2+(y-3)^2\le 4$. This leaves two cases: either $(x-2)^2+(y-3)^2=4$ or $(x-2)^2+(y-3)^2<4$. (That is, the point $(x,y)$ is either in the interior or at the boundary of the disc. Treat these separately.)
In the first case you have two conditions for two variables. Therefore there is no need to use Lagrange's method. Instead, you can solve the pair $$ \begin{cases} (x-2)^2+(y-3)^2=4\\ x^2=4y \end{cases} $$ for $(x,y)$. You will get two solutions and you can calculate the value of your function at these points.
In the second case you can use Lagrange's method with one constraint: $x^2=4y$. If this method gives you points that do not satisfy $(x-2)^2+(y-3)^2<4$, throw them away.
You will get a short list of possible minimum points. Calculate the values to see which one it is.
Some more details (which may contain errors):
In the first case, you can substitute $y=\frac14x^2$ into the first equation to get $$ (x-2)^2+(\frac14x^2-3)^2=4. $$ This has two real solutions. One ($x=2$) is neat, but the other one is unfortunately not.
In the second case, you can calculate just as you did, only with one restriction. This will lead you to $$ \begin{cases} \lambda=1\\ y=2\lambda. \end{cases} $$ Thus $y=2$ and the constraint $x^2=4y$ gives $x=\pm\sqrt{8}$.
Edit: It's not necessary to look at the problem in the interior, since it's not hard to argue that that the minimum cannot be reached there. See the comment by borisd below.