Let |z-i2z+i|=(1)/(3),z C, be the equation of a circle with center at C.

Question

Let $|\frac{\vec{z}-i}{2\vec{z}+i}|=\frac{1}{3},z\in C$, 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:

Accepted Answer

$$\begin{aligned}
& \left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3} \
& \left|\frac{\bar{z}-i}{\bar{z}+\frac{i}{2}}\right|=\frac{2}{3} \
& 3|x-i y-i|=2\left|x-i y+\frac{i}{2}\right| \
& 9\left(x^2+(y+1)^2\right)=4\left(x^2+(y-1 / 3)^2\right) \
& 9 x^2+9 y^2+18 y+9=4 x^2+4 y^2-4 y+1 \
& 5 x^2+5 y^2+22 y+8=0 \
& x^2+y^2+\frac{22}{5} y+\frac{8}{5}=0 \
& \text { centre } \Rightarrow\left(0,-\frac{11}{5}\right)
\end{aligned}$$

$$\left| {{1 \over 2}\left| {\begin{pmatrix} 0 & 0 & 1  \
   0 & { - 11/5} & 1  \
   \alpha  & 0 & 1  \

\end{pmatrix} } \right|} \right| = 11$$

$$\begin{aligned}
& \Rightarrow\left(-\frac{11}{5} \alpha\right)^2=(11 \times 2)^2 \
& \Rightarrow \alpha^2=100
\end{aligned}$$

Solutions · 1