A point P moves on the parabola y^2=4x and a point Q moves on the circle x^2+y^2-12x+35=0. The least possible value of PQ is ____.

The circle is $(x-6)^2+y^2=1$, with centre $C(6,0)$ and radius $1$. The point $C$ lies at distance more than $1$ from the parabola, so the least $PQ$ equals $\min CP-1$, since the minimising line passes through the centre. For $P(t^2,2t)$: $CP^2=(t^2-6)^2+4t^2=t^4-8t^2+36$. With $u=t^2\ge0$, $u^2-8u+36$ is minimised at $u=4$, giving $16-32+36=20$. So $\min CP=\sqrt{20}=2\sqrt5\approx4.472$. The required least distance is $2\sqrt5-1\approx3.47$. A numerical minimisation over the parabola agrees. A frequent mistake is to forget the radius subtraction and answer $4.47$.