如图 已知圆M:X^2+(y-2)^2=1,点Q是X轴上的动点,QA,QB分别切圆M与AB两点1)若|AB|=(
如图 已知圆M:X^2+(y-2)^2=1,点Q是X轴上的动点,QA,QB分别切圆M与AB两点1)若|AB|=(
(1)若|AB|=(4根号下2)除以3 ,求直线MQ的方程
(2)求证:动弦AB过定点
(1)若|AB|=(4根号下2)除以3 ,求直线MQ的方程
(2)求证:动弦AB过定点
数学人气:738 ℃时间:2019-10-18 02:28:40
优质解答
(1) 圆M的圆心M(1,2),半径r=1设AB的中点为N,Q(x0,0)则AN=(1/2)AB=2√2/3MN=√(r^2-AN^2)=1/3由MA^2=MN*MQ1=(1/3)*MQMQ=3所以(x0)^2+2^2=3^2x0=√5所以MQ的直线方程:x/√5+y/2=1(2) NQ=MQ-MN=3-1/3=8/3即MN/NQ=(1/3)...
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