Show that this locus determines a circle in C

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SUMMARY

The discussion centers on the locus of points \( z \) defined by the equation \( |(z - a)/(z - b)| = k \) where \( k \neq 1 \), which represents a circle in the Argand plane. The key insight is that the equation implies a constant angle between the vectors \( z-a \) and \( z-b \), along with a constant ratio of their lengths. This geometric interpretation confirms that the locus indeed forms a circle, with \( a \) and \( b \) as points on the circle. The center and radius of this circle can be derived from the properties of the angles and lengths involved.

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vaishakh
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a and b represent two numbers on an Argand plane, i.e. fixed. Then it is given that locus of z given by the equation
|(z - a)/(z - b)| = k where k is not 1.
Now it is given that locos of z represents a circle. I cannot understand how can this equation represent a circle.
|z -a| k|z - b| - this maks me think geometrically impossible to represent a circle.
Can anyone help me here?
The question is to find centre and radius.
 
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Sorry
|z-a| k |z -b|
it is not so
|z - a| = k|z - b| - it was a typo.
 
Geometrically, the equation is stating that the difference in angle between the two vectors is a constant, and so is their ratio of lengths. The former statement is more important. Consider a circle through the points a and b and the chord a-b. Consider z on the circle and the lines z-a and z-b. There is a simple theorem about the angle between those two lines for any point z on the circle. See http://mathworld.wolfram.com/Chord.html .
 

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