HKCEE A. Maths. (Circle)

oscar

The circles C[1]: x^2+y^2+4x-2y+1=0 and C[2]: x^2+y^2+10x+4y+F=0 intersect each other at two points P and Q, where the equation of the line PQ is x+y+3=0.
(a) Find the value of F.
(b) M is an external point of C[1] and C[2]. If M lies on the line PQ, show that the length of the tangent from M to C[1] is equal to the length of tangent from M to C[2].
===========================================
(a)我已經做完，答案是F=19.
(b)就沒有任何頭緒....

細薯

A.Maths.未教Circle....

oscar

Originally posted by smallpotato226 at 2005-3-16 19:59:
A.Maths.未教Circle....

下...咁慢ge...
你地中五既A.Maths.咪有排追？

細薯

Originally posted by oscar at 2005-3-16 08:03 PM:

下...咁慢ge...
你地中五既A.Maths.咪有排追？

唔知ar........

已教:
1)Quadratic Function
2)Inequalities
3)Binormial
4)Mathematical induction
5)Trigo.....(教緊,sum and product)
6)Striaght line

(可能串錯字)

oscar

Originally posted by smallpotato226 at 2005-3-16 20:08:

唔知ar........

已教:
1)Quadratic Function
2)Inequalities
3)Binormial
4)Mathematical induction
5)Trigo.....(教緊,sum and product)
6)Striaght line

(可能串錯字)

教完Straight Lines應該教Circle
(吾恨Geometry也...)

細薯

Originally posted by oscar at 2005-3-16 08:11 PM:

教完Straight Lines應該教Circle
(吾恨Geometry也...)

Geometry 係咩?

比超仔

不是post在自然科學區?????

抹茶可樂

C[2]-C[1]:6x+6y+F-1=0
Comparing with 6x+6y+18=0
F-1=18
F=19

Since x+y+3=0, y=-x-3
let A(a,-a-3) be an any point on PQ
Length tangent to C[1]
=√[a^2+(-a-3)^2+4a-2(-a-3)+1]
=√[a^2+a^2+6a+9+4a+2a+6+1]
=√[2a^2+12a+16]
Length tangent to C[2]
=√[a^2+(-a-3)^2+10a+4(-a-3)+19]
=√[a^2+a^2+6a+9+10a-4a-12+19]
=√[2a^2+12a+16]

∴The proof is finished

細薯

Originally posted by 比超仔 at 2005-3-16 08:14 PM:
不是post在自[color=8E ...

隨便啦
如果oscar認為好D就移

oscar

Originally posted by 落雷 at 2005-3-16 20:18:
C[2]-C[1]:6x+6y+F-1=0
Comparing with 6x+6y+18=0
F-1=18
F=19

Since x+y+3=0, y=-x-3
let A(a,-a-3) be an any point on PQ
Length tangent to C[1]
=√[a^2+(-a-3)^2+4a-2(-a-3)+1]
=√[a^2+a^2+6a+ ...

咦....係wor...
等我仲係度慢慢搵條tangent既equations tim....
我都有諗過用Length of tangent from external point，不過做唔到你咁....