F.5 Maths(Equation of Circle)

S: x² + y² - 12y + 11 = 0
Centre of S: C = (0, -(-12)/2) = (0, 6)

L: x + y - 5 = 0
Perpendicular distance between the centre C and L:
= |(0) + (6) - 5| / √(1² + 1²)
= 1/√2
= (√2)/2

2011-02-25 00:29:04 補充：
Alternative method:

S = (0, 6)
L: x + y = 5 ...... [1]
Let M = the point that L meets the perpendicular from S to L.

Slope of L = -1/1 = -1
Slope of SM = -1/(-1) = 1

Equation of SM:
(y - 6)/x = 1
x = y - 6 ...... [2]

2011-02-25 00:29:18 補充：
Put [2] into [1]:
(y - 6) + y = 5
y = 11/2

Put y = 5.5 into [2]:
x = (11/2) - 6 = -1/2

Hence, M = (-0.5, 5.5)

Perpendicular distance between the centre C and L = SM
= √{[0 + (1/2)]² + [6 - (11/2)]²}
= (√2)/2 ...... (answer)

Still have (b) part and thereafter?