已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM┃=┃PO已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM
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![已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM┃=┃PO已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM](/uploads/image/z/5185475-35-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86C%EF%BC%9Ax%5E2%2By%5E2%2B2x-4y%2B3%3D0.%E4%BB%8E%E5%9C%86%E5%A4%96%E4%B8%80%E7%82%B9P%EF%BC%88X1%2CY1%29%E5%90%91%E5%9C%86%E5%BC%95%E4%B8%80%E6%9D%A1%E5%88%87%E7%BA%BF%2C%E5%88%87%E7%82%B9%E4%B8%BAM%2CO%E4%B8%BA%E5%9D%90%E6%A0%87%E5%8E%9F%E7%82%B9%2C%E4%B8%94%E6%9C%89%E2%94%83PM%E2%94%83%3D%E2%94%83PO%E5%B7%B2%E7%9F%A5%E5%9C%86C%EF%BC%9Ax%5E2%2By%5E2%2B2x-4y%2B3%3D0.%E4%BB%8E%E5%9C%86%E5%A4%96%E4%B8%80%E7%82%B9P%EF%BC%88X1%2CY1%29%E5%90%91%E5%9C%86%E5%BC%95%E4%B8%80%E6%9D%A1%E5%88%87%E7%BA%BF%EF%BC%8C%E5%88%87%E7%82%B9%E4%B8%BAM%EF%BC%8CO%E4%B8%BA%E5%9D%90%E6%A0%87%E5%8E%9F%E7%82%B9%EF%BC%8C%E4%B8%94%E6%9C%89%E2%94%83PM)
已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM┃=┃PO已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM
已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM┃=┃PO
已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM┃=┃PO┃,求使┃PM┃最小的P点坐标。
答案是(根号2-1,2-(2又根号2))或(-根号2-1,2+2又根号2)啊~
已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM┃=┃PO已知圆C:x^2+y^2+2x-4y+3=0.从圆外一点P(X1,Y1)向圆引一条切线,切点为M,O为坐标原点,且有┃PM
圆C方程:(x+1)^2+(y-2)^2=2,所以圆心C(-1,2),R^2=2
设P点的坐标为(x,y)
则|PM|^2=|PC|^2-R^2=(x+1)^2+(y-2)^2-R^2=x^2+y^2+2x-4y+3
|PO|^2=x^2+y^2
因为|PM┃=┃PO┃,
所以x^2+y^2+2x-4y+3=x^2+y^2,得点P的轨迹方程:2x-4y+3=0
把轨迹方程代入,消去y,
得|PM|^2=(x+1)^2+(1/2x+3/4-2)^2-2
=5/4(x^2+3/5x)+9/16
=5/4(x+3/10)^2+9/20
所以当x=-3/10时,|PM|取得最小值
所以最小时的P点的坐标为(-3/10,3/5)
(-3/10,3/5).