Let $|\frac{\bar{z}-i}{2\bar{z}+i}|=\frac{1}{3}, z\in\mathbb{C}$ 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
- (1) 100
- (2) 50
- (3) $\frac{121}{25}$
- (4) $\frac{81}{25}$
Solution:
Given equation: $\left|\frac{\bar{z}-i}{2(\bar{z}+i/2)}\right| = \frac{1}{3} \implies 3|\bar{z}-i| = 2|\bar{z}+i/2|$.
Squaring both sides (Let $z = x+iy \implies \bar{z} = x-iy$):
$$9|x – i(y+1)|^2 = 4|x – i(y-1/2)|^2$$
$$9[x^2 + (y+1)^2] = 4[x^2 + (y-1/2)^2]$$
$$9(x^2 + y^2 + 2y + 1) = 4(x^2 + y^2 – y + 1/4)$$
Rearranging terms:
$$5x^2 + 5y^2 + 18y + 4y + 9 – 1 = 0$$
$$5x^2 + 5y^2 + 22y + 8 = 0$$
Dividing by 5:
$$x^2 + y^2 + \frac{22}{5}y + \frac{8}{5} = 0$$
Center of the circle $C$ is $(0, -\frac{11}{5})$.
Area of triangle with vertices $O(0,0)$, $C(0, -11/5)$, and $A(\alpha, 0)$:
$$Area = \frac{1}{2} \times \text{Base} \times \text{Height}$$
Base (on x-axis) = $|\alpha|$, Height (y-coordinate of C) = $|-\frac{11}{5}| = \frac{11}{5}$.
Given Area = 11:
$$\frac{1}{2} \times |\alpha| \times \frac{11}{5} = 11$$
$$\frac{11|\alpha|}{10} = 11 \implies |\alpha| = 10$$
$$\therefore \alpha^2 = 100$$
Ans. (1)