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问题: 已知点P(0,5)及圆C: x^2+y^2+4x-12y+24=0

已知点P(0,5)及圆C: x^2+y^2+4x-12y+24=0
(1)若直线L过点P,且与圆C的圆心相距为2,求直线L的方程;
(2)求过P点的圆C弦的重点轨迹方程.

解答:

C:
x^2 + y^2+4x-12y + 24 = 0
(x+2)^2 + (y-6)^2 = 4+36 - 24 = 16;
圆心(-2,6), 半径 4;

设L的斜率是k, 过P(0,5);
L: y = kx+5, kx-y + 5 = 0;
与圆C的圆心相距为2,
d = |-2k-1|/sqrt(1+k^2) = 2;
4k = 3, k = 3/4;
直线L的方程:
y = (3/4)x+5, 3x-4y+20 = 0;

2)
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的重点是坐标 (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 -2x -11y + 30 = 0.