Let C be the circle in the complex plane with centre z_0=12(1+3i) and radius r=1.

Question

Let $C$ be the circle in the complex plane with centre $z_0=\dfrac12(1+3i)$ and radius $r=1$. Let $z_1=1+i$ and the complex number $z_2$ be outside the circle $C$ such that $|z_1-z_0|\cdot|z_2-z_0|=1$. If $z_0, z_1, z_2$ are collinear, then the smaller value of $|z_2|^2$ is equal to

Accepted Answer

$$|z_1-z_0|=\left|1+i-\left(\frac12+\frac32i\right)\right|=\left|\frac12-\frac12i\right|=\frac1{\sqrt2}.$$

So $|z_2-z_0|=\dfrac{1}{1/\sqrt2}=\sqrt2$.

The direction from $z_0=\left(\dfrac12,\dfrac32\right)$ to $z_1=(1,1)$ has slope $\dfrac{1-3/2}{1-1/2}=-1$, corresponding to angle $135^{\circ}$. Since $z_2$ lies on the line through $z_0,z_1$ (on either side):

$$z_2 = \left(\frac12+\sqrt2\cos135^{\circ},\ \frac32+\sqrt2\sin135^{\circ}\right) \text{ or } \left(\frac12-\sqrt2\cos135^{\circ},\ \frac32-\sqrt2\sin135^{\circ}\right).$$

With $\cos135^{\circ}=-\dfrac1{\sqrt2},\sin135^{\circ}=\dfrac1{\sqrt2}$, the two candidate points are $\left(-\dfrac12,\dfrac52\right)$ and $\left(\dfrac32,\dfrac12\right)$.

$$\left|-\frac12+\frac52i\right|^2=\frac14+\frac{25}4=\frac{26}4, \qquad \left|\frac32+\frac12i\right|^2=\frac94+\frac14=\frac{10}4=\frac52.$$

The smaller value is $\dfrac52$.

Solutions · 1