Let the first circle be $C_1: x^2 + y^2 = 25$ with center $O=(0,0)$ and radius $R=5$. Let the second circle be $C_2: (x-1)^2 + (y-2)^2 = 1$ with center $P=(1,2)$ and radius $r=1$. Let $M(h,k)$ be the midpoint of a chord of $C_1$. The equation of the chord of $C_1$ whose midpoint is $(h,k)$ is given by $T=S_1'$, where $S_1'$ is the value of $S_1$ at $(h,k)$. So, $xh + yk - 25 = h^2 + k^2 - 25$, which simplifies to $xh + yk = h^2 + k^2$. This chord touches the circle $C_2$. This means the perpendicular distance from the center $P(1,2)$ of $C_2$ to the chord must be equal to its radius $r=1$. The equation of the chord can be written as $hx+ky-(h^2+k^2)=0$. Using the formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax+By+C=0$: $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$. Here, $(x_0, y_0) = (1,2)$ and the line is $hx+ky-(h^2+k^2)=0$. So, the distance is: $\frac{|h(1) + k(2) - (h^2+k^2)|}{\sqrt{h^2+k^2}} = 1$. To eliminate the absolute value and square root, square both sides of the equation: $(h + 2k - (h^2+k^2))^2 = (h^2+k^2)$. To find the locus, replace $(h,k)$ with $(x,y)$: $(x + 2y - (x^2+y^2))^2 = x^2+y^2$. This can be rearranged as $((x^2+y^2) - (x+2y))^2 = x^2+y^2$. This precisely matches option A.