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Question

Prove that the equation of circle in the z plane can be written in the from αz¯z+¯βz+β¯z+c=0 Deduce the equation of the line.

A
¯βz+β¯z+c=0
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B
¯βzβ¯z+c=0
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C
¯βz+β¯zc=0
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D
None of these
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Solution

The correct option is A ¯βz+β¯z+c=0
If z=x+iy then ¯z=xiy
x=z+¯z2,y=z¯z2i,z¯z=|z|2=x2+y2
The equation of a circle is
α(x2+y2)+2gx+2fy+c=0
α z¯z+g(z+¯z)+f(z¯z)i+c=0
or α z¯z+g(z+¯z)if(z¯z)+c=0
or α z¯z+(gif)z+(g+if)¯z+c=0
Now if g+if=β then gif=¯β
α z¯z+¯βz+β¯z+c=0
It is of the same form as given.
Deduction: Circle (1) will reduce to a straight line if α =0 in (1). Hence putting α =0 in (2) the equation of the line can be put in the form
¯βz+β¯z+c=0

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(i) If A(z1),B(z2) are two complex numbers, then distance between z1 & z2 is AB=|z1z2|
(ii) Line segment joining the points A(z1),B(z2) is divided by the point P(z) in the ratio m:n, then z=mz2+nz1m+n where m,n are real numbers.
(iii) Points z1,z2,z3 are collinear if ∣ ∣z1¯z11z2¯z21z3¯z31∣ ∣=0 and the general equation of line joining A(z1),B(z2) in non-parametric given by ∣ ∣z¯z1z1¯z11z2¯z21∣ ∣=0 and general equation of straight line is a¯z+¯az+b=0 where a is a complex number & b is real, the real slope of the line is a+¯ai(¯aa) or equal to Re(a)Img(a) and the complex slope of the line is a¯a or coefficient of ¯zcoefficient of z.The lines ¯az+a¯z+λ=0 and z¯a¯za+iλ=0 are respectively parallel and to the line a¯z+¯az+b=0
(iv) The length of from a point z1 to the line ¯az+a¯z+b=0 is given by |a¯z1+¯az1+b|2|a|.
(v) If the point A(z1),B(z2) are to be considered as foci and length of major axis is 2a then equation of ellipse is given by |zz1|+|zz2|=2a2a>|z1z2|.
If major axis is to be considered as transversal axis with length 2a such that 2a<|z1z2| The equation of hyperbola is given by zz1zz2=2a

(vi) If |zα|=r (α is complex) is a circle at α=(p,q) & radius =r
Let α,β be two fixed non zero complex numbers & 'z' a variable complex number such that the lines α¯z+¯αz+1=0 and β¯z+¯βz+1=0 are mutually perpendicular lines, then value of α¯β+β¯α equals

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