$ƒ(x,y)=\sqrt{x^2+y^2}+\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-2y+1}+\sqrt{x^2+y^2-6x-8y+25}$ QUESTION: Let $ƒ(x,y)=\sqrt{x^2+y^2}+\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-2y+1}+\sqrt{x^2+y^2-6x-8y+25}$
(A) Minimum value of $ƒ(x,y)= 5+\sqrt2$
(B) Minimum value of $ƒ(x,y)= 5-\sqrt2$
(C) Minimum value occurs of $ƒ(x,y)$ for  $x=\frac{3}{7}$ 
(D) Minimum value occurs of $ƒ(x,y)$ for  $y=\frac{4}{7}$ 
My approach , all values in the square roots needs to be positive.
${x^2+y^2\ge 0}$ hence it encloses the whole graph
${x^2+y^2-2x+1}$,${(x-1)^2+y^2\ge 0}$
$x^2+y^2-2y+1$, $ x^2+(y-1)^2\ge 0$
$x^2+y^2-6x-8y+25$, $(x-3)^2+(y-4)^2\ge 0$
Not able to approach from here

Hint. Note that the centers of the four circles are:
$$O(0,0),A(0,1),B(1,0),C(3,4)$$
For the shortest distance, the point $(x,y)$ must lie on the intersection of the line passing through $(0,0)$ and $(3,4)$ and the line passing through $(0,1)$ and $(1,0)$. Hence:
$$|OC|+|AB|=5+\sqrt2.$$
