8. $\odot P$的半径为2，圆心$P$在直线$y=2x-1$上运动. 当$\odot P$与$x$轴相切时，圆心$P$的坐标为. A

8.(1.5,2)或(-0.5,-2)

9. 如图，Rt△ABC的内切圆⊙O与两直角边AB，BC分别相切于点D，E，过$\overset{\frown}{DE}$上任一点P（不包括端点D，E）作⊙O的切线MN，与AB，BC分别交于点M，N。若⊙O的半径为4 cm，则Rt△MBN的周长为。

9.8 cm

10.如图,在△ABC中,O为AC上一点,以点O为圆心,OC的长为半径作圆,与BC相切于点C,过点A作AD⊥BO交BO的延长线于点D,且∠AOD=∠BAD.
(1)若∠AOD=60°,求∠CBD的度数;
(2)求证:AB为⊙O的切线.

10.(1)解:$\because$BC是$\odot O$的切线,C是切点,
$\therefore OC⊥ BC$,即$∠ OCB=90°$,
$\therefore ∠ BOC+∠ OBC=90°$.
$\because ∠ BOC=∠ AOD=60°$,$\therefore ∠ CBD=30°$.
(2)证明:如答图,过点O作$OE⊥ AB$于点E.

由题意可知$∠ CBD=∠ OAD$.$\because AD⊥ BO$,$\therefore ∠ D=90°$,
即$∠ OAD+∠ AOD=90°$,$∠ ABD+∠ BAD=90°$.
$\because ∠ AOD=∠ BAD$,$\therefore ∠ OAD=∠ ABD$,
$\therefore ∠ ABD=∠ CBD$,即BD是$∠ ABC$的平分线.
$\because OC⊥ BC$,$OE⊥ AB$,$\therefore OC=OE$.
$\because OC$是$\odot O$的半径,$\therefore$点O到AB的距离OE等于$\odot O$的半径,$\therefore AB$是$\odot O$的切线.

11.如图，AB是$\odot O$的直径，AD和BC分别切$\odot O$于A，B两点，CD与$\odot O$有公共点E，且$AD=DE$.
(1)求证:CD是$\odot O$的切线；
(2)若$AB=12$，$BC=4$，求AD的长.

11.(1)证明:如答图,连接OD,OE.
$\because AD$切$\odot O$于点A,AB是$\odot O$的直径,$\therefore ∠ DAB=90°$.
$\because AD=DE$,$OA=OE$,$OD=OD$,
$\therefore △ ADO≌△ EDO(\mathrm{SSS})$,
$\therefore ∠ OED=∠ OAD=90°$,
$\therefore CD$是$\odot O$的切线.

(2)解:如答图,过点C作$CH⊥ AD$于点H.
$\because AB$是$\odot O$的直径,AD和BC分别切$\odot O$于A,B两点,
$\therefore ∠ DAB=∠ ABC=∠ CHA=90°$,
$\therefore$四边形ABCH是矩形,
$\therefore CH=AB=12$,$AH=BC=4$.
$\because CD$是$\odot O$的切线,$\therefore AD=DE$,$CE=BC$,
$\therefore DH=AD-BC=AD-4$,$CD=AD+4$.
$\because CH^2+DH^2=CD^2$,
$\therefore 12^2+(AD-4)^2=(AD+4)^2$,解得$AD=9$.