1、课题:课题:7.37.3圆圆【知识点梳理知识点梳理】:建立适当的直角坐标系(建系)建立适当的直角坐标系(建系)222rbyax0422FED2,2ED2422FEDr0422FED)(sincos是参数是参数ryrx)(sincos是参数是参数rbyraxrPC 22020rbyaxrPC 22020rbyaxrPC 22020rbyaxrd 0)(联联立立方方程程rd 0)(联联立立方方程程rd 0)(联联立立方方程程rRPQrR0)(联联立立方方程程0)(联联立立方方程程rRPQrRPQ0)(联联立立方方程程rRPQrRPQ222dr 200ryyxxCMNdr【例题精讲例题精讲】:解:解
2、:则由角平分线的性质,得则由角平分线的性质,得:222923325.14292xyxy整理,得整理,得:2340,3270.xyxy解:解:代入抛物线方程,得:代入抛物线方程,得:122122xy解:解:0b1a或或12ba解:解:由题意,得:由题意,得:rbaaarbar1)2()1(1122222852183rba02543yx0432yxx或或解:解:223222baba032abbabb30或或所以,所求切线方程是:所以,所求切线方程是:解:解:由题意,得:由题意,得:3),0,2(rC半径半径圆心圆心rCOdrCO347,34722yx,22132213,13xysin3cos32
3、yx则则)4sin(6262,62证明:证明:由题意,得:由题意,得:0),(,0),(002001yxfyxf且且Ryxfyxf,0),(),(002001解:解:0)1(3313122yxyx即:即:因为方程是直线方程,因为方程是直线方程,31-049yx所以,公共弦所在的直线方程所以,公共弦所在的直线方程 是:是:23:得得解:解:07204yxyx联立联立13yx55rAC又又.内内在圆在圆CA)1,2(ACn0)1()3(2yxl的方程:的方程:.052 yx即:即:解:解:解:解:解:解:rd 则则374374k证明:证明:22117kxx(定定值值)7ANAM由题意,得:由题意,得:1:kxyl11322kk)1,()1,(2211yxyxANAM)1()1(2121yyxx2122121)1(xxkkxkxxx1,1,2211yxANyxAM则则2121yyxxONOM则则1281)1(422121kkkyyxxONOM1 k解:解:由题意,得:由题意,得:0)1(2811622kk-)(374374k-,17221kxx112121kxkxyy1)(21212xxkxxk11)1(417222kkkkk