Question

Let \( \frac{\overline{z} - i}{z - i} = \frac{1}{3}, \, z \in \mathbb{C} \), 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 \( (\alpha, 0) \), is 11 square units, then \( \alpha^2 \) equals:

• A. \( 100 \)
• B. \( 50 \)
• C. \( 121 \)
• D. \( \frac{81}{25} \)

Hint:

For complex geometry problems involving circles and distances in the complex plane, converting the given equation to a standard form and using properties like the modulus and area of triangles can simplify the calculations.

Answer 1

The Correct Option is A

\(\left| \frac{z-i}{2z+i} \right| = \frac{1}{3}\)

\(\left| \frac{z-i}{z+1} \right| = \frac{2}{3}\)

3\(|x - iy - i| = 2 |x - iy + \frac{i}{2}|\)

\(9(x^{2} + (y+1)^{2}) = 4(x^{2} + (y - \frac{1}{3})^{2})\)

\(9x^{2} + 9y^{2} + 18y + 9 = 4x^{2} + 4y^{2} - 4y + 1\)

\(5x^{2} + 5y^{2} + 22y + 8 = 0\)

\(x^{2} + y^{2} + \frac{22y}{5} + \frac{8}{5} = 0\)

centre \((0, -\frac{11}{5})\)

\[ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \quad \left| \frac{10}{5} \right| = 11 \]

\(\alpha = 1\)

\(\Rightarrow \left( \frac{-11}{5} \alpha \right)^{2} = (11 \times 2)^{2}\)

\(\Rightarrow \alpha^{2} = 100\)