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Question

Intercept made by the circle z¯¯¯z+¯¯¯az+a¯¯¯z+r=0 on the real axis on complex plane is

A
(a+¯¯¯a)r
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B
(a+¯¯¯a)22r
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C
(a+¯¯¯a)24r
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D
(a+¯¯¯a)28r
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Solution

The correct option is C (a+¯¯¯a)24r
Points where the circle cuts the xaxis z=¯¯¯z.
Hence subsituting z=¯¯¯z in the equation of circle, we get
z2+¯¯¯az+az+r=0
z2+(a+¯¯¯a)z+r=0
AB=|z1z2|
(where A and B are points of intersectoin of circle with x axis)
=(z1+z2)24z1z2
=(a+¯¯¯a)24r

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