1.設(shè)直線L為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
直線L為x-2y+5=0,或2x-y-5=0
2.
(1)設(shè)線段AB中點坐標(biāo)M(x,y),圓上一點(x0,y0)
則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(復(fù)雜)