Circle Geometry

Using coordinate geometry, let P(0, 0) be the origin, the centre of the circle is O (a, r) with radius r.
So, the equation of the circle is
Circle : (x – a)² +(y – r)² = r² . .. (1)
Let the equation of straight line passes through P and cut the circle at C, B be
(equation of PCB) : y = mx . . . . . (2)
ie. the coordinates of B and C be (b, mb) and (c,mc) respectively.
The equation of OP is
(equation of PEO) : y = rx/a . . . . (3)
So, the coordinates are : O(a, r), A(a, 2r), B(b,mb), C(c, mc), D(a, 0), P(0, 0)
The equation of AC is :
(y – 2r)/(x – a) = (mc – 2r)/(c – a) . . . . . . .. . (two-points form)
==> (mc - 2r)x + (a – c)y + c(2r – ma) = 0 . . .(4)
To find the coordinates of point E, sub (2) into(4), gets,
a(mc – 2r)x + r(a – c)x + ac(2r – ma) = 0
==> x = ac(2r – ma)/(ar + cr – mac)
From (2), y = cr(2r – ma)/(ar + cr – mac)
ie. the coordinates of E is (ac(2r – ma)/(ar + cr –mac), cr(2r – ma)/(ar + cr – mac))
Slope of ED is : [cr(2r – ma)/(ar + cr – mac) –0]/[ac(2r – ma)/(ar + cr – mac) – a]
ie. slope of ED : c(2r – ma)/(ac - a²) .. . . . (5)
And slope of AB : (2r –mb)/(a – b) . . . . . . (6)
Solve (1), (3), get :
(1 + m²)x² -2(a + mr)x + a² = 0
As points B, C are the solution of (1), (3),therefore
b + c = 2(a + mr)/(1 + m²) .. . . . (7)(sum of roots)
bc = a²/(1 + m²) . . . . . . . . . . . . . (8)(product of roots)
slope of ED – slope of AB (ie. (5) – (6))
= c(2r – ma)/(ac - a²) –(2r – mb)/(a – b)
= [c(2r – ma)(a – b) – (ac - a²)(2r– mb)]/[(ac - a²)(a– b)]
Consider the numerator,
c(2r – ma)(a – b) – (ac - a²)(2r– mb)
= 2acr – 2bcr - ma²c +mabc – 2acr + mabc + 2a²r -ma²b
= -2bcr - ma²c +2mabc – 2acr + 2a²r -ma²b= 2mabc - ma²(b +c) – 2bcr + 2a²r
= 2ma*a²/(1 + m²) - ma²*2(a + mr)/(1 + m²) –2r*a²/(1+ m²) +2a²r .. . (from (7), (8))
= (2ma³ - 2ma³ - 2m²a²r –2a²r + 2a²r + 2m²a²r)/(1 + m²)
= 0Therefore,
slope of ED – slope of AB = 0
ie. slope of ED = slope of AB ==> ED//ABDone.

原來應該是這樣的，我計咗好長時間了。

2012-07-13 10:58:28 補充：
請問版主, 只用 geometry 計倒嗎?

Using co- ordinate geometry acceptable ?

2012-07-10 22:15:59 補充：
Let D be (0,0) and radius of circle be 1, so center of circle O is (0,1) and A is (0,2). So equation of circle is x^2 + y^2 - 2y = 0.
For simplicity, let B be (1,1) and P be ( - 4,0). So equation of line PB is 5y = x + 4.
Solving these 2 equations, we can find that the co-ordinates of C is ( -12/13, 8/13).

2012-07-10 22:24:20 補充：
Equation of line OP is y = 1/4( x + 4), that is 4y = x + 4
Equation of line AC is (y - 2)/x = (8/13 - 2)/(-12/13) = 18/12 = 3/2, that is 2y = 3x + 4.
Solving these 2 equations, we can find that the co-ordinates of E is (-4/5, 4/5).

2012-07-10 22:26:22 補充：
Slope of AB = (1 - 2)/(1 - 0) = - 1.
Slope of ED = (4/5 - 0)/(- 4/5 - 0) = - 1
so AB//ED.