Question

已知圆C：x2+y2-2x+4y-4=0.问是否存在斜率为1的直线l,使l被圆C截得弦AB,以AB为直径的圆经过原点.

Answer 1

let A(x1,y1) ,B(x2,y2)

C：x^2+y^2-2x+4y-4=0 (1)

斜率为1的直线l : y=x+c (2)

sub (2) into (1)

x^2+(x+c)^2-2x+4(x+c)-4=0

2x^2 + (2+2c)x + c^2+4c-4 =0

x1+x2 = -(1+c)

x1x2 = (c^2+4c-4)/2

similarly

(y-c)^2+y^2-2(y-c)+4y-4=0

2y^2 +(2-2c)y +c^2+2c-4 =0

y1+y2 = -(1-c)

y1y2 = (c^2+2c-4)/2

|AB|^2 = (x1-x2)^2 +(y1-y2)^2

=(x1+x2)^2-4x1x2 + (y1+y2)^2 - 4y1y2

= (1+c)^2 - 2(c^2+4c-4) + (1-c)^2 -2(c^2+2c-4)

= 18-12c-2c^2

|AB|^2 = 4r^2

r^2 = (18-12c-2c^2)/4

O' = mid point of AB

=( (x1+x2)/2 , (y1+y2)/2 )

=( -(1+c)/2 , -(1-c)/2)

equation of cicle with centre O' and radius r : C1

[x+ (1+c)/2]^2 +[y+(1-c)/2 ]^2 = (18-12c-2c^2)/4

orgin (0,0)

[ (1+c)/2]^2 +[(1-c)/2 ]^2 = (18-12c-2c^2)/4

1+c^2 =9-6c-c^2

c^2+3c-4=0

c=-4 or 1

=> 存在斜率为1的直线l