From a variable point P on the line x+y=10, two tangents are drawn to the circle x^2+y^2=25, and the chord of contact is AB. The minimum possible…

For $P(h,k)$ with $h+k=10$, the chord of contact is $hx+ky=25$. Substituting $k=10-h$ gives $h(x-y)+10y-25=0$ for every $h$, so the chord always passes through the fixed point $F$ given by $x=y$, $10y=25$, i.e. $F\left(\tfrac52,\tfrac52\right)$. A chord through $F$ has length $2\sqrt{25-d^{2}}$, where $d$ is its distance from the centre $O$, and $d\le OF=\tfrac{5\sqrt2}{2}$. The length is least when $d=OF$, i.e. when $AB\perp OF$. Then $AB=2\sqrt{25-\tfrac{25}{2}}=2\cdot\tfrac{5}{\sqrt2}=5\sqrt2$. This chord belongs to $P=(5,5)$, the point of the line nearest $O$ (indeed $AB=2\cdot5\sqrt{1-25/50}=5\sqrt2\approx7.07$). Option (B) $10$ is the diameter, attained only if the line passed through $O$; (C) and (D) result from using $d=0$ incorrectly or dropping the factor 2.