Let there be two parabolas: $$y^2+2kx=k^2 \tag{1}$$ $$x^2+2ky=k^2 \tag{2}$$
What will be the point of intersection of these two curves? The general form of intersection of such two curves would lead to solving a quartic equation(a lengthy solution), which I want to avoid, since there is clear symmetry in the two parabolas, as visible after plotting them, and also evident from the Wolfram's Solution. So, I'm trying to find the solution by exploiting the symmetry (if possible), and not by going through the quartic equation.
Along the lines of this post, I added eqns (1) and (2): $$y^2+x^2+2ky+2kx=2k^2$$ After completing the square, $$(y+k)^2+(x+k)^2=(2k)^2\tag{3}$$
Let $k>0$, then the circle in (3) has centre at $O'=(-k,-k),$ and radius $2k$. For $k=6$, the parabolas and the circle look like this: 
Clearly, the first intersection point $P(x,y)$ has $x>0,y>0$. Drawing a radius from centre of circle $O'$ to point $P(x,y)$, $O'P=2k$. Let origin be $O(0,0)$. Then $O'O= \sqrt2 k$. As $O'P=O'O+OP,$ $ OP=2k-\sqrt 2 k= \sqrt2 k(\sqrt2-1)$. Also, $OP=\sqrt{x^2+y^2},$ so $\sqrt{x^2+y^2}= \sqrt2 k(\sqrt 2 -1)$
Now I only need to put $x=y$ to obtain $x=y=k(\sqrt 2-1)$, which is indeed one of the solution given by Wolfram Alpha (for $k>0$). However, I can't think of how to state $x=y$; I assumed this after seeing the solution by WA. But, if we let $x=y$ in (1) and (2), then it reduces to solving a quadratic and all solutions can be obtained as $x=y=\pm \sqrt2 |k|-k$.
By (3), the circle has centre at $(-k,-k)$, which means the circle remains the same even if we interchange the X and Y axis. And since the intersection points lie on the circle, so $x=y$ for intersection point(s) $P(x,y)$. Is this reasoning correct? It feels more like an assumption, but it gives the answer, so it is true, but how can it be proven? Or is there any alternative for finding solution?
Since one curve is the reflection of the other across the line $y = x$, there are two possible kinds of intersections:
To find points of the first kind, replace $y$ be $x$ in one of the equations. (Choose either equation; the result is the same.) Now you just have to solve a quadratic equation in $x$.
The rest of the solution is to show that there are no other solutions. Graphically, it's obvious, but I don't propose that as a proof.
From $x^2+2ky=y^2+2kx$ we can derive
$$ (x + y)(x - y) = 2k(x - y), $$
so either $x - y = 0$ (the first kind of solution) or $x + y = 2k$. Substituting $2k - x$ for $y$ in Equation $(1)$, we find that $x = k = 0$, hence also $y = 0$, so this is actually not a solution of the second kind but just the first kind of solution for the special case where $k = 0$.
In fact, the solution of $x^2+2kx=k^2$ along with the assertion $y = x$ finds both of the actual solutions.