If the center and radius of the circle | (z - 2)/(z - 3) | = 2 are respectively (α,β) and γ, then 3(α+β+γ) is equal to :

Solution

Maths sir · 4 steps

Put z = x + iy, square both sides, and the modulus equation turns into an ordinary circle.

1. Remove the fraction. The condition is |z-2| = 2|z-3|. Put z = x + iy and square both sides, so no square roots are left.

   (x-2)² + y² = 4[(x-3)² + y²]

2. Expand and tidy. This gives x² - 4x + 4 + y² = 4x² - 24x + 36 + 4y², that is 3x² + 3y² - 20x + 32 = 0. Divide by 3 to get x² + y² - 20/3x + 32/3 = 0.
3. Read off the circle. The centre is (10/3, 0), so α = 10/3 and β = 0. The radius is γ = √(100/9 - 32/3) = √(4/9) = 2/3.
4. So. α + β + γ = 10/3 + 0 + 2/3 = 4, and 3(α + β + γ) = 12, option (a).

The idea to keep. |z - z₁| = k|z - z₂| with k ≠ 1 is always a circle. Square both sides with z = x + iy, make the coefficient of x² equal to 1, then read off the centre and radius.

Answer from the NTA final answer key. Solution by PrepWiser.