For concentric circles, it is not difficult to find out that, if expressed in the Argand diagram, the equation is given by:
|z - z0| = r where z0 is the complex no. representing the common centre and r is any non-negative real values.
How about the expressions for the family of:
1. passing through the points of intersection between a circle and a line; and
2. passing through the points of intersection between 2 circles.
Family of circles in complex
2009-07-28 2:22 am
回答 (4)
2009-08-05 2:58 pm
✔ 最佳答案
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my wisdom of pure maths
2009-08-06 07:11:31 補充：
nelsonywm2000，飛天魏國大將軍張遼想要的其實是形如x² + y² + Dx + Ey + F + k(Ax + By + C) = 0和形如x² + y² + D1x + E1y + F1 + k(x² + y² + D2x + E2y + F2) = 0般的答案，並不是想你玩雕琢constant的遊戲。
2009-08-06 07:20:27 補充：
chowchihong2004，你的答案是錯的。因為在aZ + b (barZ) + c = 0中，只有當b = bar a的時候才是直線，b ≠ bar a的時候肯定不是直線，而是變成另一種曲線。
2009-08-05 3:08 am
1. | z-d|^2 - r^2 + k( aZ + b (barZ) + c) = 0 ,
2. |z-a|^2 - r^2 + k( |z-b|^2 - R^2|) = 0
* abcd are complex numbers and k , r, R are real numbers
Moreover If a and b are the points in the Argand plane
the family of the circles passing through them is :
lm[(z-a){ (bar)(z-b)} = k |z-a||z-b|
2. |z-a|^2 - r^2 + k( |z-b|^2 - R^2|) = 0
* abcd are complex numbers and k , r, R are real numbers
Moreover If a and b are the points in the Argand plane
the family of the circles passing through them is :
lm[(z-a){ (bar)(z-b)} = k |z-a||z-b|
2009-08-02 3:53 am
(1)Equation of a straight line: | z – a | = | z – b | equivalent to say all points that are equidistant from a & b.
Equation of the circle | z – c | = r
Refer to the diagram, http://img220.imageshack.us/img220/2979/complexcircle.jpg
The centre of a member circle is u = c + k(b – a), that is the line joining c and u is parallel to the line joining a and b.
Mid point of a and b is (a+b)/2
p2 – x2 = q2 – y2 ...(1)
r2 – x2 = R2 – y2 …(2)
(2) – (1) =>
r2 – p2 = R2 – q2
R2 = r2 – p2 + q2
p = | (a+b)/2 – c |
q = | (a+b)/2 – c – k(b-a) |
General equation is | z – u | = R
where u = c + k(b – a)
R2 = r2 – | (a+b)/2 – c |2 + | (a+b)/2 – c – k(b-a) |2
(2)
Circle 1 : | z – a | = p
zz’ – za’ – z’a + aa’ = p2
Circle 2 : | z – b | = q
zz’ – zb’ – z’b + bb’ = q2
Family of circles:
zz’ – za’ – z’a + aa’ + k(zz’ – zb’ – z’b + bb’) = p2 + kq2
(1 + k) zz’ – z(a’ + kb’) – z’(a + kb) + aa’ + kbb’ = p2 + kq2
zz’ – z(a’ + kb’) / (1 + k) – z’(a + kb) / (1 + k) = (p2 + kq2 – aa’ – kbb’) / (1 + k)
zz’ – z(a’ + kb’) / (1 + k) – z’(a + kb) / (1 + k) + (a’ + kb’)(a + kb) / (1 + k)2 = (p2 + kq2 – aa’ – kbb’)/ (1 + k) + (a’ + kb’)(a + kb) / (1 + k)2
Let u = (a + kb) / (1 + k)
zz’ – zu’ – z’u + uu’ = (p2 + kq2 – aa’ – kbb’)/ (1 + k) + (a’ + kb’)(a + kb) / (1 + k)2
| z – u |2 = (p2 + kq2 – aa’ – kbb’)/ (1 + k) + (a’ + kb’)(a + kb) / (1 + k)2
| z – u |2 = (p2 + kq2)/ (1 + k) – k(b’ – a’)(b –a) / (1 + k)2
| z – u |2 = (p2 + kq2)/ (1 + k) – k| (b – a)/(1 + k) |2
| z – u | = R
Centres of the circles = (a + kb) / (1 + k)
R = √ [(p2 + kq2)/ (1 + k) – k| (b – a)/(1 + k) |2]
Equation of the circle | z – c | = r
Refer to the diagram, http://img220.imageshack.us/img220/2979/complexcircle.jpg
The centre of a member circle is u = c + k(b – a), that is the line joining c and u is parallel to the line joining a and b.
Mid point of a and b is (a+b)/2
p2 – x2 = q2 – y2 ...(1)
r2 – x2 = R2 – y2 …(2)
(2) – (1) =>
r2 – p2 = R2 – q2
R2 = r2 – p2 + q2
p = | (a+b)/2 – c |
q = | (a+b)/2 – c – k(b-a) |
General equation is | z – u | = R
where u = c + k(b – a)
R2 = r2 – | (a+b)/2 – c |2 + | (a+b)/2 – c – k(b-a) |2
(2)
Circle 1 : | z – a | = p
zz’ – za’ – z’a + aa’ = p2
Circle 2 : | z – b | = q
zz’ – zb’ – z’b + bb’ = q2
Family of circles:
zz’ – za’ – z’a + aa’ + k(zz’ – zb’ – z’b + bb’) = p2 + kq2
(1 + k) zz’ – z(a’ + kb’) – z’(a + kb) + aa’ + kbb’ = p2 + kq2
zz’ – z(a’ + kb’) / (1 + k) – z’(a + kb) / (1 + k) = (p2 + kq2 – aa’ – kbb’) / (1 + k)
zz’ – z(a’ + kb’) / (1 + k) – z’(a + kb) / (1 + k) + (a’ + kb’)(a + kb) / (1 + k)2 = (p2 + kq2 – aa’ – kbb’)/ (1 + k) + (a’ + kb’)(a + kb) / (1 + k)2
Let u = (a + kb) / (1 + k)
zz’ – zu’ – z’u + uu’ = (p2 + kq2 – aa’ – kbb’)/ (1 + k) + (a’ + kb’)(a + kb) / (1 + k)2
| z – u |2 = (p2 + kq2 – aa’ – kbb’)/ (1 + k) + (a’ + kb’)(a + kb) / (1 + k)2
| z – u |2 = (p2 + kq2)/ (1 + k) – k(b’ – a’)(b –a) / (1 + k)2
| z – u |2 = (p2 + kq2)/ (1 + k) – k| (b – a)/(1 + k) |2
| z – u | = R
Centres of the circles = (a + kb) / (1 + k)
R = √ [(p2 + kq2)/ (1 + k) – k| (b – a)/(1 + k) |2]
2009-07-29 7:52 am
1. [ |z - z1| ^2 - r1^2 ] + k[ |z - z2| ^2 - |z - z3| ^2 ] = 0
2. [ |z - z1| ^2 - r1^2 ] + k[ |z - z2| ^2 - r2^2 ] = 0
2. [ |z - z1| ^2 - r1^2 ] + k[ |z - z2| ^2 - r2^2 ] = 0
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