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Sum
If `|(z - 2)/(z + 2)| = pi/6`, then the locus of z is ______.
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Solution
If `|(z - 2)/(z + 2)| = pi/6`, then the locus of z is circle.
Explanation:
Given that: `|(z - 2)/(z + 2)| = pi/6`
Let z = x + iy
⇒ `|(x + iy - 2)/(x + iy + 2)| = pi/6`
⇒ `|((x - 2) + iy)/((x + 2) + iy)| = pi/6`
⇒ `6|(x - 2) + iy| = pi|(x + 2) + iy|`
⇒ `6sqrt((x - 2)^2 + y^2) = pisqrt((x + 2)^2 + y^2)`
⇒ `36[x^2 + 4 - 4x + y^2] = pi^2[x^2 + 4 + 4x + y^2]`
⇒ 36x2 + 144 – 144x + 36y2 = π2x2 + 4π2 + 4π2x + π2y2
⇒ (36 – π2)x2 + (36 – π2)y2 – (144 + 4π2)x + 144 – 4π2 = 0
Which represents are equation of a circle.
Concept: Quadratic Equations
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