Let $x,y$ be real numbers such that $x^2+y^2-4x-6y+13=0$. Find the value of $x^y$.

Step 1: Re-arrange the given equation to group the variables:

Step 2: Complete the square for both $x$ and $y$:

• $x^2-4x=(x-2{)}^2-4$

• $y^2-6y=(y-3{)}^2-9$

Substitute these back into the equation:

Step 3: Since both term squares $(x-2{)}^2\ge 0$ and $(y-3{)}^2\ge 0$ are non-negative for real numbers, their sum can only be $0$ if each term is individually $0$:

Step 4: Calculate $x^y$:

So the value is $8$.