中四數學一問,勁人請入

Let O be centre of circle and the shortest distance to PQ be x, so shortest distance to RS = 6 - x.
OP = OR = radius of circle = r.
By Pythagoras theorem,
5^2 + x^2 = r^2 = 7^2 + (6 - x)^2
25 + x^2 = 49 + 36 + x^2 - 12x
12x = 60
x = 5 cm.

1a)設圓心為O ,半徑為r ,PQ中點為 m , RS中點為 M :

情況一 : 當 M , m 在 O 之下(或上), Om > 6

情況二 :當 O 位於 M , m 之間 , Om < 6,因此應取情況二。

MO + mO = 6

(r^2 - (14/2)^2) + (r^2 - (10/2)^2) = 6

(r^2 - 49) + (r^2 - 25) + 2 [(r^2 - 49)(r^2 - 25)] = 36

r^2 + [(r^2 - 49)(r^2 - 25)] = 55

2009-10-13 18:09:43 補充：
(r^2 - 49)(r^2 - 25) = (55 - r^2)^2

r^4 - 49r^2 - 25r^2 + 1225 = r^4 - 110r^2 + 3025

36r^2 = 1800

r^2 = 50

r = 50

圓心至弦PQ的最短距離 = mO = (50 - (10/2)^2)

= 5 cm

圖在哪?