求最值

$已知P(x_0,y_0) 是圆C:x^2+y^2-2x-2y+1=0上任意一点，则\cfrac{y_0+1}{x_0-3}的最小值为(\qquad )$
$A.\cfrac{4+\sqrt{7}}{3}\quad B.\cfrac{-4-\sqrt{7}}{3}\quad C.\cfrac{4-\sqrt{7}}{3}\quad D.\cfrac{-4+\sqrt{7}}{3}\quad$
$令\cfrac{y_0+1}{x_0-3}=k\Rightarrow y_0+1=k(x_0-3),即转换成直线与圆C相交时，直线斜率k的最小值$
$\begin{cases} y=k(x-3)-1\\x^2+y^2-2x-2y+1=0\end{cases}\Rightarrow x^2+[k(x-3)-1]^2-2x-2[k(x-3)-1]+1=0$
$\Delta =0,解得k 值。计算量巨大$
$我们应用圆心(1,1)到切线的距离不大于半径,(x-1)^2+(y-1)^2=1,d=\cfrac{|k(1-3)-1-1|}{\sqrt{1+k^2}}\le 1$
$4(k+1)^2\le 1+k^2\Rightarrow 3k^2+8k+3\le 0\Rightarrow \cfrac{-4-\sqrt{7}}{3}\le k\le \cfrac{-4+\sqrt{7}}{3},C选$

$设实数x,y满足x^2+(y-1)^2=1,则\cfrac{y-x}{x-2}的最大值为(\qquad)$
$A.\cfrac{1}{6}\quad B.\cfrac{1}{4}\quad C.\cfrac{1}{3}\quad D.\cfrac{1}{2}\quad$
$分离常数\cfrac{y-x}{x-2}=-1+\cfrac{y-2}{x-2},故转换成上题一样。先令\cfrac{y-2}{x-2}=k,k(x-2)-y+2=0$
$圆心(0,1)到切线y-2=k(x-2)的距离不大于半径1,d=\cfrac{|k(0-2)-1+2|}{\sqrt{1+k^2}}\le 1$
$(2k+1)^\le 1+k^2\Rightarrow 3k^2-4k\le 0\quad \Rightarrow 0\le k\le \cfrac{4}{3}, \cfrac{4}{3}-1=C$