已知圆C:x²+y²+x-6y+m=0与直线l:x+2y-3=0相交于P,Q两点,O为原点,若向量OP·向量OQ=0.(1)求实数m的值;(2)若R(x,y)为圆C上一点,求x+y-5/6m的最大值与最小值.
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![已知圆C:x²+y²+x-6y+m=0与直线l:x+2y-3=0相交于P,Q两点,O为原点,若向量OP·向量OQ=0.(1)求实数m的值;(2)若R(x,y)为圆C上一点,求x+y-5/6m的最大值与最小值.](/uploads/image/z/4033922-50-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86C%3Ax%26%23178%3B%2By%26%23178%3B%2Bx-6y%2Bm%3D0%E4%B8%8E%E7%9B%B4%E7%BA%BFl%3Ax%2B2y-3%3D0%E7%9B%B8%E4%BA%A4%E4%BA%8EP%2CQ%E4%B8%A4%E7%82%B9%2CO%E4%B8%BA%E5%8E%9F%E7%82%B9%2C%E8%8B%A5%E5%90%91%E9%87%8FOP%C2%B7%E5%90%91%E9%87%8FOQ%3D0.%281%29%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%80%BC%3B%282%29%E8%8B%A5R%28x%2Cy%29%E4%B8%BA%E5%9C%86C%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E6%B1%82x%2By-5%2F6m%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%8E%E6%9C%80%E5%B0%8F%E5%80%BC.)
已知圆C:x²+y²+x-6y+m=0与直线l:x+2y-3=0相交于P,Q两点,O为原点,若向量OP·向量OQ=0.(1)求实数m的值;(2)若R(x,y)为圆C上一点,求x+y-5/6m的最大值与最小值.
已知圆C:x²+y²+x-6y+m=0与直线l:x+2y-3=0相交于P,Q两点,O为原点,若向量OP·向量OQ=0.
(1)求实数m的值;
(2)若R(x,y)为圆C上一点,求x+y-5/6m的最大值与最小值.
已知圆C:x²+y²+x-6y+m=0与直线l:x+2y-3=0相交于P,Q两点,O为原点,若向量OP·向量OQ=0.(1)求实数m的值;(2)若R(x,y)为圆C上一点,求x+y-5/6m的最大值与最小值.
(1)将 x= 3-2y 代入圆的方程得 (3-2y)^2+y^2+(3-2y)-6y+m=0 ,
化简得 5y^2-20y+12+m=0 ,
设P(x1,y1),Q(x2,y2),
则 y1+y2= 4 ,y1*y2=(12+m)/5 ,
因此 x1*x2=(3-2y1)(3-2y2)=9-6(y1+y2)+4y1y2=4/5*(12+m)-15 ,
由于 OP*OQ=0 ,因此 x1x2+y1y2=0 ,
即 12+m-15=0 ,
解得 m=3 .
(2)由(1)得圆的方程为 (x+1/2)^2+(y-3)^2=25/4 ,
因此圆心为(-1/2 ,3),半径 r=5/2 ,
令 t=x+y-5/6*m ,则直线方程化为 x+y-t-5/2=0 ,
由已知,该直线与圆有公共点,所以圆心到直线的距离不超过半径 ,
即 |-1/2+3-t-5/2|/√2<=5/2 ,
化简得 |t|<=5√2/2 ,
解得 -5√2/2<= t <= 5√2/2 ,
所以,所求最大值为 5√2/2 ,最小值为 -5√2/2 .