圆C:x^2+y^2+4x-12y+24=0 求证:所有过P(0,5)点的圆C的弦的中点的轨迹是一个圆

圆C:x^2+y^2+4x-12y+24=0
过P(0,5)点的圆C的弦AB的中点M(x,y)
2x=xA+xB,2y=yA+yB
k(AB)=(yA-yB)/(xA-xB)=(y-5)/x
x^2+y^2+4x-12y+24=0
(xA)^2+(yA)^2+4xA-12yA+24=0......(1)
(xB)^2+(yB)^2+4xB-12yB+24=0......(2)
(1)-(2):
(xA)^2+(yA)^2+4xA-12yA+24-[(xB)^2+(yB)^2+4xB-12yB+24]=0
[(xA)^2-(xB)^2]+[(yA)^2-(yB)^2]+4(xA-xB)-12(yA-yB)=0
(xA+xB)*(xA-xB)+(yA+yB)*(yA-yB)+4(xA-xB)-12(yA-yB)=0
xA≠xB
(xA+xB)+(yA+yB)*[(yA-yB)/(xA-xB)]+4-12(yA-yB)/(xA-xB)=0
2x+2y*(y-5)/x+4-12(y-5)/x=0
(x+1)^2+(y-5.5)^2=1.25
所有过P(0,5)点的圆C的弦的中点的轨迹是一个圆

P点代入圆方程：
2^2+1^2 = 5 < 16,
P点在圆内，设过P点的直线斜率是k，
y = kx +5,
代入圆方程：
(1+k^2)x^2-2(k-2)x-11 = 0;
设AB是直线与圆的交点，坐标为： (xa,ya), (xb,yb);
xa, xb是方程的两个根；

xa + xb = 2(k-2)/(1+k^2);
ya + yb = k(xa+xb)+10;

弦AB的中点是坐标 (x,y);
x = (xa+xb)/2 = (k-2)/(1+k^2);
y = (ya+yb)/2 = k*(k-2)/(1+k^2) + 5;

(y -5)/x = k;
消去k,
x^2 + y^2 -