Let $\left|\frac{\bar z-i}{2\bar z+i}\right|=\frac13$, $z\in C$, 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:

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $9|z+i|^2=|2z-i|^2$ gives $9\left(x^2+(y+1)^2\right)=4x^2+(2y-1)^2$, i.e. $x^2+y^2+\frac{22}5y+\frac85=0$. Centre $C\left(0,-\frac{11}5\right)$. Area of the triangle $=\frac12\cdot\frac{11}5\cdot|\alpha|=11$, so $|\alpha|=10$ and $\alpha^2=100$. Answer (2).

Q. Let $\left|\frac{\bar z-i}{2\bar z+i}\right|=\frac13$, $z\in C$, 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:

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?