From a point P moving on the line x+y=6, tangents are drawn to the circle x^2+y^2=9. The chord of contact of these tangents always passes through a…

Let $P=(h,6-h)$. Its chord of contact with $x^2+y^2=9$ is $hx+(6-h)y=9$, i.e. $h(x-y)+(6y-9)=0$. This holds for every $h$ only if $x=y$ and $6y=9$, so the fixed point is $F=\left(\tfrac32,\tfrac32\right)$. Every chord of contact is a chord of the circle through $F$. For a chord at distance $d$ from the centre the length is $2\sqrt{9-d^2}$, and $d\le OF$ because the chord passes through $F$. Hence the length is least when $d$ is greatest, i.e. when the chord is perpendicular to $OF$ and $d=OF=\tfrac{3}{\sqrt2}$. The minimum length is $2\sqrt{9-\tfrac92}=2\sqrt{\tfrac92}=3\sqrt2$. This chord is the chord of contact of the point $P=(3,3)$ on the line (the foot of the perpendicular from $O$), so it is attained. Option $6$ would be the diameter (the maximum), not the minimum.