Question

高中数学必修2题一.直线L经过P（5,5）,且和圆C：X2（平方）+Y2（平方）=25相交,截得弦长4倍根号5,求L的方

Answer 1

1.设直线L为y=kx-5k+5,y^2=k^2x^2+(10k-10k^2)x+25k^2-50k+25

(1+k^2)x^2+(10k-10k^2)x+25k^2-50k=0

X1+x2=(10k^2-10k)/(1+k^2),x1x2=(25k^2-50k)/(1+k^2)

(X1-x2)^2=(X1+x2)^2-4x1x2=(10k^2-10k)^2/(1+k^2)^2-4(25k^2-50k)/(1+k^2)

=[(10k(k-1))^2-100k(k-2)(1+k^2)]/(1+k^2)^2

(y1-y2)^2=k^2(X1-x2)^2

(X1-x2)^2+(y1-y2)^2=(1+k^2)(X1-x2)^2

=[(10k(k-1))^2-100k(k-2)(1+k^2)]/(1+k^2)

[(10k(k-1))^2-100k(k-2)(1+k^2)]/(1+k^2)=80

整理得2k^2-5k+2=0,解得k1=1/2,k2=2

直线L为x-2y+5=0,或2x-y-5=0

2.

(1)设线段AB中点坐标M(x,y),圆上一点(x0,y0)

则x=(x0+1)/2,x0=2x-1；y=(y0+3)/2,y0=2y-3

(2x)^2+(2y-3)^2=4,整理x^2+(y-3/2)^2=1

(2)待解(复杂)