coordinate geometry (only 2)

27.
(a)
Let x² + y² + Dx + Ey + F = 0 be theequation of the circle.

(0, 0) lies on the circle :
(0)² + (0)² + D(0) + E(0) + F = 0
F = 0
Hence, the equation o the circle becomes : x² + y² + Dx + Ey = 0

(0, 5) lies on the circle :
(0)² + (5)² + D(0) + E(5) = 0
5E = -25
E = -5

(12, 0) lies on the circle :
(12)² + (0)² + D(12) + Ey + F = 0
12D = -144
D = -12

Hence, the equation of the circle :
x² + y² - 12x - 5y = 0

(b)
Radius of the circle
= √[(D/2)² + (E/2)² - F]
= √[(-5/2)² + (-12/2)²]
= 13/2


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28.
Set 1 : a = 1, b = 2, c = 5, d = 2

Set 2 : a = 3, b = 0, c = 3, d = 4

2013-08-20 22:29:23 補充：
27.(a)
The equation of the circle : x² + y² - 12x - 5y = 0
Hence, the coordinates of the centre = (12/2, 5/2) = (6, 5/2)

2013-08-20 22:29:43 補充：
27.(a)
The equation of the circle : x² + y² - 12x - 5y = 0
Hence, the coordinates of the centre = (12/2, 5/2) = (6, 5/2)

27a) coordinate of center= ((12+0)/2, (5+0)/2)= (6,5/2)
b)radius =√[(6-0)^2+(5/2 -5)^2]=13/2


28) any integers a,b,c,d satisfying a+c=6, b+d=4

set1: a=1,b=1,c=5,d=3
set2: a=2,,b=2,c=4,d=2