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1. 如图,在平面直角坐标系中,$\odot P的圆心坐标为(6,a)(a＞5)$,半径为5,函数$y= x的图像被圆所截得的弦AB$的长为8,则$a$的值为(
A.6
B.$6+\sqrt{2}$
C.$3\sqrt{2}$
D.$6+3\sqrt{2}$
D
)A.6
B.$6+\sqrt{2}$
C.$3\sqrt{2}$
D.$6+3\sqrt{2}$
答案：
D
2. 如图,在平面直角坐标系中,$\odot O经过点(0,10)$,直线$y= kx+2k-4与\odot O交于B$,$C$两点,则弦$BC$的最小值是
$8\sqrt{5}$
。
答案：
$8\sqrt{5}$
3. 如图,$AB$,$AC是\odot O$的两条弦,$M是\overset{\frown}{AB}$的中点,$N是\overset{\frown}{AC}$的中点,弦$MN分别交AB$,$AC于点P$,$D$。
(1) 求证：$AP= AD$；
(2) 连接$PO$,若$AP= 3$,$OP= \sqrt{10}$,$\odot O$的半径为5,求$MP$的长。
(1) 求证：$AP= AD$；
(2) 连接$PO$,若$AP= 3$,$OP= \sqrt{10}$,$\odot O$的半径为5,求$MP$的长。
答案：
(1) 证明: 如答图, 连接 $AM$, $AN$.
由题意知 $\overset{\frown}{AM} = \overset{\frown}{BM}$, $\overset{\frown}{AN} = \overset{\frown}{CN}$,
$\therefore \angle BAM = \angle ANM$, $\angle AMN = \angle CAN$,
$\because \angle APD = \angle AMN + \angle BAM$, $\angle ADP = \angle CAN + \angle ANM$, $\therefore \angle APD = \angle ADP$, $\therefore AP = AD$.
(2) 解: 连接 $OM$ 交 $AB$ 于点 $E$, 连接 $AO$. 设 $PE = x$,
$\because \overset{\frown}{AM} = \overset{\frown}{BM}$, $\therefore OM \perp AB$, $\therefore \angle AEO = 90^{\circ}$,
$\because OE^{2} = OA^{2} - AE^{2} = OP^{2} - PE^{2}$,
$\therefore 5^{2} - (x + 3)^{2} = (\sqrt{10})^{2} - x^{2}$, $\therefore x = 1$,
$\therefore AE = 4$, $OE = 3$, $ME = 2$,
$\therefore MP = \sqrt{EM^{2} + PE^{2}} = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$.
(1) 证明: 如答图, 连接 $AM$, $AN$.
由题意知 $\overset{\frown}{AM} = \overset{\frown}{BM}$, $\overset{\frown}{AN} = \overset{\frown}{CN}$,
$\therefore \angle BAM = \angle ANM$, $\angle AMN = \angle CAN$,
$\because \angle APD = \angle AMN + \angle BAM$, $\angle ADP = \angle CAN + \angle ANM$, $\therefore \angle APD = \angle ADP$, $\therefore AP = AD$.
(2) 解: 连接 $OM$ 交 $AB$ 于点 $E$, 连接 $AO$. 设 $PE = x$,
$\because \overset{\frown}{AM} = \overset{\frown}{BM}$, $\therefore OM \perp AB$, $\therefore \angle AEO = 90^{\circ}$,
$\because OE^{2} = OA^{2} - AE^{2} = OP^{2} - PE^{2}$,
$\therefore 5^{2} - (x + 3)^{2} = (\sqrt{10})^{2} - x^{2}$, $\therefore x = 1$,
$\therefore AE = 4$, $OE = 3$, $ME = 2$,
$\therefore MP = \sqrt{EM^{2} + PE^{2}} = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$.
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