求过直线2X+Y+4=0和圆X^2+Y^2+2X-4Y+1=0的交点,且经过(2,-1)的圆的方程

Answer 1

2x+y+4=0
y=-4-2x
M:x^2+y^2+2x-4y+1=0
(x+1)^2+(y-2)^2=4
M(-1,2)
x^2+(-4-2x)^2+2x-4*(-4-2x)+1=0
5x^2+26x+33=0
弦AB的中点P
xA+xB=-26/5,xA*xB=33/5
(xA-xB)^2=(xA+xB)^2-4xA* xB=
(yA-yB)^2=4(xA-xB)^2
AB^2=(xA-xB)^2+(yA-yB)^2=5(xA-xB)^2
(AB/2)^2=AB^2/4
L:2x+y+4=0,k(L)=k(AB)=-2
k(MP)=1/2
MP:y-2=(1/2)*(x+1)
x=2y-5
C(2c-5,c),已知D(2,-1)
d^2=|2*(2c-5)+c+4|^2/25
r^2=CD^2=(2c-7)^2+(c+1)^2
(AB/2)^2+d^2=r^2
c=
方法正确，请自己计算。