Adding these equations we have : $x+y=x^2+ax+b+y^2+cx+d$. $x^2+ax+cx-x+y^2-y+b+d=0$, $x^2+(a+c-1)x+y^2-y+b+d=0$. $x^2+(a+c-1)x+\frac{(a+c-1)^2}{4}+y^2-y+\frac{1}{4}+b+d=\frac{(a+c-1)^2}{4}+\frac{1}{4}$. $(x+\frac{a+c-1}{2})^2+(y-\frac{1}{2})^2=\frac{(a+c-1)^2+1}{4}-b-d$. And this is the equation of a circle.Q.E.D. Ok

Pirkuliyev Ronsen: your solution is incorrect, but it's only a small mistake. Notice that the second equation contains $cy$ and not $cx$, and notice that the equation you get at the end is not necessarily the equation of a circle: therefore the $RHS$ should be positive, since it's the radius squared. Otherwise there are no real solutions.

Non-algebraic solutions can be found here https://math.stackexchange.com/questions/134366/intersection-of-two-parabolae

The problem is equivalent to the following. Quote: Let $A$, $B$, $C$, $D$ be concyclic points. Then the two parabolas through $A$, $B$, $C$, $D$ [if they exist] have perpendicular axes. Let $\ell_{\infty}$ be the line at infinity. Define an involution $h$ on $\ell_{\infty}$ by sending each $P \in \ell_{\infty}$ to the second intersection of conic $ABCDP$ and $\ell_{\infty}$. Since $ABCD$ is cyclic, the bisectors of $\angle (\overline{AB}, \overline{DC})$ and $\angle (\overline{AD}, \overline{BC})$ are perpendicular to each other. Thus, the involution $h$ is described by reflection over one of these two lines. In particular, $h$ has two fixed points, corresponding to perpendicular directions. Since a parabola is a conic tangent to the line at infinity, these two fixed points give the two parabolas possible, and they have perpendicular axes.