Complex Number solutions for a Circle

When you graph the solutions in "the plane", you are restricting yourself to look at solutions to the equation where both $x$ and $y$ are real. You could, for example, restrict further to only allow $x$ and $y$ to be rational numbers, and think about how those points fit in with all the real solutions.

To think geometrically about non-real complex solutions, you will need more (real) dimensions! You could restrict yourself, as it sounds like you are doing in the question, to just solutions where $x$ is real and $y$ is allowed to be complex. Then you will need another dimension/direction for the imaginary part of $y$. You could graph this in a "$z$" direction, so that solutions where $x>5$ not lie in the plane but above/below it. You will find that for $|x|>5$ the solutions will be points where $x^2 - z^2 = 25$ so that if you just look at the $(x,z)$ plane the solution set will look like a hyperbola.

Probably the most interesting thing to look at is when you allow both $x$ and $y$ to be complex... but graphing this would require more dimensions.

Let me denote by $A$ the circle you consider: $$ A=\{(x,y)\in\mathbb R^2;x^2+y^2=25\}. $$ This is the usual real circle and this is what a circle in the plane means unless something different is specifically explained. This set consists of all pairs $(x,y)$ of numbers $x$ and $y$ which are real numbers and satisfy the equation $x^2+y^2=25$.

I stress that it is part of the very definition of $A$ that the two numbers are real. You can ask whether a point $(x,y)$ can be in $A$ if $x=6$. As you found out, if $x=6$ and the equation $x^2+y^2=25$ is satisfied, then $y=\pm i\sqrt{11}$. But since $y$ is not in $\mathbb R$, the point $(x,y)$ is not in $A$. What fails is not the equation $x^2+y^2=25$ but the condition that $x\in\mathbb R$ and $y\in\mathbb R$. You could say that you have found a solution of the equation which is not in the plane $\mathbb R^2$.

You can also consider a different set: $$ B=\{(x,y)\in\mathbb C^2;x^2+y^2=25\}. $$ This set looks otherwise similar to $A$, but the two numbers $x$ and $y$ can now be complex. You can call $B$ a complexification of $A$. Since $\mathbb R\subset\mathbb C$, we have $A\subset B$. But there are points in $B$ which are not in $A$. For example, the point $(6,-i\sqrt{11})$ is in $B$ — both numbers are complex and the equation is satisfied — but not in $A$ as discussed above. However, the set $B$ is not what is usually called a circle. It is a sort of an extension of the circle, but in a weird way.

If you identify $\mathbb C$ with $\mathbb R^2$, then $B$ is a two-dimensional surface in $\mathbb R^4$, whereas $A$ is a one-dimensional curve in $\mathbb R^2$. Intersecting the surface with a suitable two-dimensional plane in $\mathbb R^4$ gives a circle ($B\cap\mathbb R^2=A$).