The circles C1: x^2+y^2=25 and C2: (x-7)^2+(y-1)^2=25 intersect at two points A and B.

The centres are $O_1(0,0)$ and $O_2(7,1)$, so $O_1O_2^2=50=r_1^2+r_2^2$; hence the circles are orthogonal (A true). Subtracting the equations gives $14x+2y-50=0$, i.e. the radical axis is $7x+y=25$, not $7x+y=50$ (C false). Solving with $x^2+y^2=25$ gives $A(3,4)$ and $B(4,-3)$, so $AB=\sqrt{1+49}=5\sqrt2$, not $5$ (B false). The quadrilateral $O_1AO_2B$ is a kite with perpendicular diagonals $O_1O_2=5\sqrt2$ and $AB=5\sqrt2$, so its area is $\tfrac12\cdot5\sqrt2\cdot5\sqrt2=25$ (it is in fact a square of side $5$), not $50$ (D false).