$ \sqrt{x^2+y^2-4x+2y+5}+ \sqrt{x^2+y^2-20x-10y+125}= \sqrt{(x-2)^2+(y+1)^2}+ \sqrt{(x-10)^2+(y-5)^2} \geq (Minkovski)$ $ \geq \sqrt{ \Big((x-2)+(10-x) \Big)^2+ \Big((y+1)+(5-y) \Big)^2}=10.$ Thus $\sqrt{x^2+y^2-4x+2y+5}+\sqrt{x^2+y^2-20x-10y+125} \geq 10 $ so the equality must occur. Hence $ (x-2)(5-y)=(y+1)(10-x) \Leftrightarrow 3x-4y=10. $ From $3x-4y=10$ and $3x+4y=26$ we get $\boxed{(x,y)=(6,2)}.$

We have $y=\dfrac{26-3x}{4}$ and $\sqrt{(x-2)^2+(y+1)^2}+\sqrt{(x-10)^2+(y-5)^2}=10$ On the $Oxy$ plane, consider points $A(x,y);B(2,-1);C(10,5)\Rightarrow AB+AC=BC\Rightarrow A\in BC$ Note that the line passing through $B$ and $C$ is $d: y= \dfrac{3}{4}x-\dfrac{5}{2},$ and $x\in d$ $\Rightarrow \dfrac{26-3x}{4}=\dfrac{3}{4}x-\dfrac{5}{2}\Rightarrow x=6\Rightarrow y=2$ So the only root of the given system of equations is $\boxed{(x,y)=(6,2)}$

How is this from 2016???

Minkowski inequality kills this problem