Let | z-i2 z+i|=13, zC, define a circle C with centre (0,).

Question

Let $\left|\dfrac{\bar z-i}{2\bar z+i}\right|=\dfrac13,\ z\in\mathbb{C}$, define a circle $C$ with centre $(0,\alpha)$. If the area of the triangle whose vertices are at the origin, $(0,\alpha)$, and a related point is 11 square units, then $25\alpha^2$ equals:

Accepted Answer

Let $z=x+iy$. From $\left|\dfrac{\bar z-i}{2\bar z+i}\right|=\dfrac13$:
$$9\big(x^2+(y+1)^2\big)=4\left(x^2+\left(y-\tfrac13\right)^2\right)$$
$$9x^2+9y^2+18y+9=4x^2+4y^2-\tfrac{8}{3}y+\tfrac19$$
$$5x^2+5y^2+22y+8=0 \implies x^2+y^2+\frac{22}{5}y+\frac{8}{5}=0.$$

This is a circle with centre $\left(0,-\dfrac{11}{5}\right)$ and radius$^2=\left(\dfrac{11}{5}\right)^2-\dfrac85=\dfrac{121-40}{25}=\dfrac{81}{25}$.

*(The final triangle-area step that connects this circle to the value $25\alpha^2$ was too fragmented in the source OCR to fully re-derive symbolically; the source's final numeric answer, 100, is preserved.)*

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