（5分）$(x-1)^{2}+y^{2}=25$的圆心与抛物线$y^{2}=2px(p > 0)$的焦点$F$重合，两曲线与第一象限交于点$P$，则原点到直线$PF$的距离为 _____．
答案：$\dfrac{4}{5}$．
分析：推导出$F(1,0)$，从而$p=2$，进而$y^{2}=4x$，联立$\left\{\begin{array}{l}{(x-1)^{2}+{y}^{2}=25}\\ {{y}^{2}=4x}\end{array}\right.$，得$P(4,4)$，求出直线$PF$的方程为$4x-3y-4=0$，由此能求出原点到直线$PF$的距离．
解：$\because (x-1)^{2}+y^{2}=25$的圆心与抛物线$y^{2}=2px(p > 0)$的焦点$F$重合，
$\therefore F(1,0)$，$\therefore p=2$，
$\therefore y^{2}=4x$，
联立$\left\{\begin{array}{l}{(x-1)^{2}+{y}^{2}=25}\\ {{y}^{2}=4x}\end{array}\right.$，得$\left\{\begin{array}{l}{x=4}\\ {y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\ {y=-4}\end{array}\right.$，
$\because$两曲线与第一象限交于点$P$，$\therefore P(4,4)$，
$\therefore$直线$PF$的方程为$\dfrac{y-4}{x-4}=\dfrac{0-4}{1-4}=\dfrac{4}{3}$，即$4x-3y-4=0$，
$\therefore$原点到直线$PF$的距离为$d=\dfrac{\vert -4\vert }{\sqrt{{4}^{2}+(-3)^{2}}}=\dfrac{4}{5}$．
故答案为：$\dfrac{4}{5}$．
点评：本题考查圆心坐标、抛物线方程、直线方程、点到直线距离等基础知识，考查运算求解能力，是中档题．