I used the Pythagorean theorem to represent the distance $\sqrt{(x-1)^2 + (y-2)^2}$.
Then I used the Langragian to obtain the following equation:
$\mathcal{L}(x,y) = \sqrt{(x-1)^2 + (y-2)^2} - \lambda(x^2 + y^2)$
After the evaluation of the partial derivatives and solving the equations I found: $x = \frac{1}{2}y$, by substitution in the circle equation I found $\frac{5}{4}y^2 = 80$ which means $y = \pm 8$ and by using substitution again we find $x = \pm4$. So we have 4 possibilities for finding the min and max: $(4,8) (-4,8) (4,-8) (-4,-8)$
I literally substituted each one and found that $(-4,-8)$ is max and $(4,8)$ is min. My first question is: is this correct?
My second question is what happened to lambda? How exactly is lambda useful here (also in general?) I understand why it exists (because the two gradient vectors are collinear but not necessarily equal) but I don't understand how it can be used! I know this is a weird question but I am confused.
You are over-complicating this. Your answers are correct but you don't need Lagrange multipliers.
If a point $p$ is inside a circle $C$ then the shortest and farthest distances to $C$ lie along a diameter through $p$.
To see why this is true, we look at two cases.
If $p$ is the center of the circle, then every point on $C$ is equidistant from $p$ so it doesn't matter.
If $p$ is not the center of the circle then there will be two distances along a diameter through $p$: $r_m$ and $r_M$ where $r_m\lt r_M$.
Construct two circles $C_m$ and $C_M$ of radii $r_m$ and $r_M$, respectively, centered at $p$. Both of these circles will be tangent to $C$.
The circle of radius $r_m$ will be inside of $C$. Since the point a distance of $r_m$ from $p$ is the only point on $C$ and every other point on $C_m$ lies inside of $C$, every other point that is a distance of $r_m$ away is closer to all other points of $C$. Therefore, that point is the closest.
The circle $C_M$ of radius $r_M$ will contain the circle $C$. Therefore, the point on $C$ that is a distance of $r_M$ from $p$ will be the only point on $C$ and since $C$ is contained in $C_M$, every other point on $C$ will be closer. Therefore, the point on $C_M$ a distance of $r_M$ away from $p$ will be the farthest.
For your special case, the diameter through $(1,2)$ is the line $y=2x$. This gives $$x^2+(2x)^2=80\implies 5x^2=80\implies x^2=16\implies x=\pm 4$$
This gives the two points $(4,8)$ and $(-4,-8)$ as you found using Lagrange multipliers.