Question

8£®ÒÑÖªÔ²M¹ýE£¨1£¬-1£©£¬F£¨-1£¬1£©Á½µã£¬ÇÒÔ²ÐÄÔÚx+y-2=0ÉÏ£¬
£¨1£©ÇóÔ²MµÄ·½³Ì£»
£¨2£©Èô¹ýµã£¨-2£¬2£©µÄÖ±Ïß±»Ô²MËù½ØµÃµÃÏÒ³¤Îª$2\sqrt{3}$£¬Çó¸ÃÖ±Ïߵķ½³Ì£»
£¨3£©ÈôPΪֱÏß3x+4y+8=0ÉϵĶ¯µã£¬¹ýP×öÔ²MµÄÇÐÏߣ¬ÇеãΪA£¬B£¬Çóµ±$\overrightarrow{|{PA}|}$µÄ×îСֵ£¬²¢Çó´Ëʱ$\overrightarrow{PA}•\overrightarrow{PB}$µÄÖµ£®

Answer 1

·ÖÎö £¨1£©Éè³öÔ²µÄ±ê×¼·½³Ì£¬ÀûÓÃÔ²M¹ýE£¨1£¬-1£©£¬F£¨-1£¬1£©Á½µã£¬ÇÒÔ²ÐÄÔÚx+y-2=0ÉÏ£¬½¨Á¢·½³Ì×飬¼´¿ÉÇóÔ²MµÄ·½³Ì£»
£¨2£©Çó³öÔ²Ðĵ½Ö±ÏߵľàÀëΪ1£¬·ÖÀàÌÖÂÛ£¬¼´¿ÉÇó¸ÃÖ±Ïߵķ½³Ì£»
£¨3£©ÓÉÌâÒ⣬$\overrightarrow{|{PA}|}$×îСʱ£¬MP´¹Ö±ÓÚÖ±Ïߣ¬´Ëʱ|MP|=$\frac{|3+4+8|}{\sqrt{9+16}}$=3£¬$\overrightarrow{|{PA}|}$=$\sqrt{5}$£¬¼´¿ÉÇó´Ëʱ$\overrightarrow{PA}•\overrightarrow{PB}$µÄÖµ£®

½â´ð ½â£º£¨1£©ÉèÔ²MµÄ·½³ÌΪ£º£¨x-a£©²+£¨y-b£©²=r²£¨r£¾0£©£¬
¸ù¾ÝÌâÒâµÃ$\left\{\begin{array}{l}{£¨1-a£©^{2}+£¨-1-b£©^{2}={r}^{2}}\\{£¨-1-a£©^{2}+£¨1-b£©^{2}={r}^{2}}\\{a+b-2=0}\end{array}\right.$£¬½âµÃ£ºa=b=1£¬r=2£¬
¹ÊËùÇóÔ²MµÄ·½³ÌΪ£º£¨x-1£©²+£¨y-1£©²=4£®
£¨2£©Ô²µÄ°ë¾¶Îª2£¬Ö±Ïß±»Ô²MËù½ØµÃµÃÏÒ³¤Îª$2\sqrt{3}$£¬¡àÔ²Ðĵ½Ö±ÏߵľàÀëΪ1£®
Ö±ÏßµÄбÂʲ»´æÔÚʱ£¬x=-2£¬Ô²Ðĵ½Ö±ÏߵľàÀëΪ3£¬²»·ûºÏÌâÒ⣻
Ö±ÏßµÄбÂÊ´æÔÚʱ£¬Éè·½³ÌΪy-2=k£¨x+2£©£¬¼´kx-y+2k+2=0£¬
¡àÔ²Ðĵ½Ö±ÏߵľàÀëΪ$\frac{|3k+1|}{\sqrt{{k}^{2}+1}}$=1£¬¡àk=0»ò-$\frac{3}{4}$£¬
¡àÖ±Ïߵķ½³ÌΪy=2»ò3x+4y-2=0£»
£¨3£©ÓÉÌâÒ⣬$\overrightarrow{|{PA}|}$×îСʱ£¬MP´¹Ö±ÓÚÖ±Ïߣ¬´Ëʱ|MP|=$\frac{|3+4+8|}{\sqrt{9+16}}$=3£¬¡à$\overrightarrow{|{PA}|}$=$\sqrt{5}$£¬
sin¡ÏAPM=$\frac{2}{3}$£¬cos¡ÏAPM=$\frac{\sqrt{5}}{3}$£¬¡àcos¡ÏAPB=2cos²¡ÏAPM-1=$\frac{1}{9}$£¬
¡à$\overrightarrow{PA}•\overrightarrow{PB}$=$\frac{5}{9}$£®

µãÆÀ ±¾Ì⿼²éÖ±ÏßÓëÔ²µÄ±ê×¼·½³Ì£¬¿¼²éѧÉú·ÖÎö½â¾öÎÊÌâµÄÄÜÁ¦£¬È·¶¨Ô²ÐÄÓë°ë¾¶Êǹؼü£¬ÊôÓÚÖеµÌ⣮