If the locus of zC such that Re(z-12z+i)+Re( z-12 z-i)=2 is a circle of radius r and centr…

Question

If the locus of $z\in\mathbb{C}$ such that $\operatorname{Re}\left(\dfrac{z-1}{2z+i}\right)+\operatorname{Re}\left(\dfrac{\bar z-1}{2\bar z-i}\right)=2$ is a circle of radius $r$ and centre $(a,b)$, then $15ab/r^2$ is equal to:

Accepted Answer

The condition simplifies (since the two terms are conjugates of each other) to $2\operatorname{Re}\left(\dfrac{z-1}{2z+i}\right)=2$, i.e. $\operatorname{Re}\left(\dfrac{z-1}{2z+i}\right)=1$.

Let $z=x+iy$:
$$\operatorname{Re}\left[\frac{(x-1)+iy}{2x+i(2y+1)}\right]=1 \implies \frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1)^2}=1.$$

$$2x^2-2x+2y^2+y = 4x^2+4y^2+4y+1 \implies 2x^2+2y^2+3y+2x+1=0$$
$$\implies x^2+y^2+x+\frac32y+\frac12=0.$$

Centre: $\left(-\dfrac12,-\dfrac34\right)$, and $r^2=\dfrac14+\dfrac9{16}-\dfrac12=\dfrac{5}{16}$.

$$a=-\frac12,\ b=-\frac34,\ r^2=\frac5{16}.$$

$$\frac{15ab}{r^2}=15\cdot\left(-\frac12\right)\left(-\frac34\right)\cdot\frac{16}{5}=15\cdot\frac38\cdot\frac{16}{5}=18.$$

Solutions · 1