Maths problem: how to do (b), thanks.?

圓心 (4,6) 與直線 y = 1 的距離是 |6-1| = 5.
所以圓 C 方程式是
    (x-4)^2 + (y-6)^2 = 5^2

L 斜率 -1/2, x-截距 k, 方程式
    y = (-1/2)(x-k)
交圓 C 於 A, B, 所以
    (x-4)^2+ [-1/2(x-k)-6]^2 = 25
      = x^2-8x+16 +x^2/4 -2(x/2)(k/2-6)+(k/2-6)^2
∴ (5/4)x^2 - (k/2+2)x + (k/2-6)^2-9 = 0
   x = {(2+k/2) ± √[125-(k-16)^2]}/(5/2)

∴ A, B 之中點座標:
   x = (2+k/2)/(5/2) = (4+k)/5
   y = (-1/2)[(4+k)/5-k] = 2(k-1)/5


(計算正確與否請自己檢查.)