Let |(z̄-i)/(2 z̄+i)|=1/3, z ∈ C, be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points (0,0), C and (α, 0) is 11 square units, then α² equals:

Solution

Maths sir · 4 steps

Clear the modulus by squaring both sides, find the circle's centre, then the area gives the base α.

1. Set up the circle. Put z = x + iy, so z̄ = x - iy. Cross-multiply to get 3|z̄ - i| = |2z̄ + i|, then square both sides.

   9(x² + (y+1)²) = 4x² + (2y-1)²

2. Simplify. Expanding gives 9x² + 9y² + 18y + 9 = 4x² + 4y² - 4y + 1, so 5x² + 5y² + 22y + 8 = 0. Dividing by 5: x² + y² + 22/5y + 8/5 = 0. The centre is C = (0, -11/5) and the radius is 9/5, so the circle is real.
3. Use the area. The base runs from (0,0) to (α, 0) on the x-axis, so its length is |α|. The height is the distance of C from the x-axis, which is 11/5. So 1/2 × |α| × 11/5 = 11, which gives |α| = 10.
4. So. α² = 100, which is option (b).

The idea to keep. For |z - a| = k|z - b| with k ≠ 1, square both sides to get a circle and read off its centre. If the triangle has its base on the x-axis, its height is just the centre's distance from that axis.

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