 # Complex locus of a circle

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 Complex locus of a circle Yue Kwok Choy (1) It is easy to show that |z – z1| = a , where z1C, aR form a circle with centre P1(z1) and radius a , using an Argand Diagram. (2) By putting z = x + yi and z1 = x1 + y1i , we can transform the equation to well known Cartesian form : (x – x1)2 + (y – y1)2 = a2 . The equation, in fact, is a circle with centre (x1, y1) and radius a in the rectangular plane. (3) Squaring the equation of circle in (1), we get |z – z1|2 = a2  We get another form of circle: , aC, cR . Here in order not to get an imaginary or degenerate circle . (4) Putting z = x + yi, z1 = x1 + y1i in (3) gives back the Cartesian form of the circle. (5) Putting z = r(cos  + i sin ) , z1 = (cos  + i sin ) , ( ,  are constants) in (3) : we get the polar form of a circle : , with centre = (, ) and radius = . (6) , 0 <  <  gives an arc and not a circle. As in the figure, the locus gives an arc of the circle standing on the chord with end points z1 and z2 such that P1PP2 =  is subtended by the chord at points on the arc, using the s in the same segment theorem. (7) Putting z = x + yi , z1 = x1 + y1i , z2 = x2 + y2i in (6), we have :     (8) The last equation in (7) is a homogenous equation of degree 2 , also (a) coeff. of x2 –term = coeff. of y2 term and (b) there is no xy-term, It therefore gives a complete circle and not an arc. The "problematic" step in (7) marked by " " changes the arc into a circle . (9) The locus of P in (7) represents : (a) when  = 0 , the whole line P1P2 with the line segment P1P2 removed. (b) when  =  , the line segment P1P2 . (c) when 0 <  < , an arc of a circle, terminating at P1 and P2 (and excluding these points) (d) when  = /2 , a semicircle and the supplementary semicircle is given by  = 3/2 . (10) or  +  , 0 <  <  gives a complete circle with P1 and P2 removed. You may investigate the following loci : (a) or  –  , 0 <  <  . (b) , 0 <  <  . (11) where z1 , z2 C, k > 0 , k  1 gives a circle (excluding points P1(z1), P2(z2) ) Note : When k = 1, the locus is the perpendicular bisector of the line joining P1(z1) and P2(z2) . Proof :     Comparing this with that given in (3), we get a circle with centre and radius a, where on simplification (exercise) (12) and if we take P(z) a variable point and P1(z1) and P2(z2) , we have P1P = k P2P . This then reduces to a well-known geometry problem : The Circle of Apollonius: Given two fixed points P1 and P2, the locus of point P such that the ratio of P1P to P2P is constant , k, is a circle. The Circle of Apollonius is not discussed here. Interested readers may consult web-sites such as: http://jwilson.coe.uga.edu/emt725/Apollonius/Cir.html If we know that the locus is a circle, then finding the centre and radius is easier. As in the diagram, C is the centre and AB is the diameter of the circle. Then A and B divide P1P2 internally and externally : P1 A : AP2 = k : 1 P1 B : BP2 = –k : 1  By section formula: A represents B represents  The centre of the circle represents and the radius = . Dostları ilə paylaş:

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