Question

高中数学图中这题第二问怎么求啊

Answer 1

圆心A(-1,2) ，直线l1:x+2y+7=0 相切， 过点B(0,5)的动直线l与圆A相交于M，N两点
(1)
A(-1,2)到 直线l1:x+2y+7=0 的距离 =r
r= |-1+2(2)+7|/√(1^2+2^2) = 2√5
圆的方程 : (x+1)^2 +(y-2)^2 =20
(2)
B(0,5) 的直线方程
y-5=mx
mx-y+5 =0 (1)
圆的方程 : (x+1)^2 +(y-2)^2 =20 (2)
M(x1,y1) ,N(x2,y2)
sub (1) into (2)
(x+1)^2 +(y-2)^2 =20
(x+1)^2 +(mx+5-2)^2 =20
(x+1)^2 +(mx+3)^2 =20
(1+m^2)x^2 +(2+6m)x -11 =0
x1+x2 = -(2+6m)/(1+m^2)
(x1+x2)/2 = 0
-(1+3m)/(1+m^2) =0
m= -1/3
若以MN 为直径的圆经过原点时，直线l的斜率 =m = -1/3

追问

圆经过原点而不是圆心是原点，为什么x1+x2=0

追答

(2)
不好意思，这样才对！
B(0,5) 的直线方程
y-5=mx
mx-y+5 =0 (1)
圆的方程 : (x+1)^2 +(y-2)^2 =20 (2)
M(x1,y1) ,N(x2,y2)
sub (1) into (2)
(x+1)^2 +(y-2)^2 =20
(x+1)^2 +(mx+5-2)^2 =20
(x+1)^2 +(mx+3)^2 =20
(1+m^2)x^2 +(2+6m)x -10 =0
x1.x2 = -10/(1+m^2) (3)
同样地
sub (1) into (2)
(x+1)^2 +(y-2)^2 =20
[(y-5)/m+1]^2 +(y-2)^2 =20
(y-5 +m)^2 +m^2.(y-2)^2 =20m^2
(1+m^2)y^2 + [ 2(m-5) -4m^2]y + (m-5)^2 +4m^2 -20m^2 =0
(1+m^2)y^2 + [ 2(m-5) -4m^2]y -15m^2 -10m +25 =0
y1.y2 = (-15m^2 -10m +25)/(1+m^2) (4)
∠MON = 90°
=>

OM的斜率. ON的斜率 =-1
(y1/x1)(y2/x2) = -1
[ (-15m^2 -10m +25)/(1+m^2) ]/[ -10/(1+m^2) ] =-1
-15m^2 -10m +25 = 10
15m^2+10m-15 =0
3m^2+2m -3 =0
m= (-1+√10)/3 or (-1-√10)/3
若以MN 为直径的圆经过原点时，直线l的斜率 =m = (-1+√10)/3 or (-1-√10)/3