Question

If $\left|z+8\right|+\left|z-8\right|=16$, where $z$ is a complex number, then the point $z$ will lie on

• A. a circle
• B. an ellipse
• C. a straight line
• D. None of the above

Answer 1

The correct option is C

a straight line

Explanation for the correct option.

Find the locus of point $z$.

Let $z=x+iy$, then the equation $\left|z+8\right|+\left|z-8\right|=16$ can be written as:

$$\begin{gathered} \left|z+8\right|+\left|z-8\right|=16 \\ \Rightarrow \left|x+iy+8\right|=16-\left|x+iy-8\right| \\ \Rightarrow \left|{\left(x+8\right)}^2+iy\right|=16-\left|\left(x-8\right)+iy\right| \\ \Rightarrow \sqrt{{\left(x+8\right)}^2+y^2}=16-\sqrt{{\left(x-8\right)}^2+y^2} \end{gathered}$$

Now, square both sides.

$$\begin{gathered} {\left(\sqrt{{\left(x+8\right)}^2+y^2}\right)}^2={\left(16-\sqrt{{\left(x-8\right)}^2+y^2}\right)}^2 \\ \Rightarrow x^2+16x+64+y^2=256+x^2-16x+64+y^2-32\sqrt{{\left(x-8\right)}^2+y^2} \\ \Rightarrow -256+32x=-32\sqrt{{\left(x-8\right)}^2+y^2} \\ \Rightarrow 8-x=\sqrt{{\left(x-8\right)}^2+y^2} \end{gathered}$$

Now again square both sides.

$$\begin{gathered} {\left(8-x\right)}^2={\left(\sqrt{{\left(x-8\right)}^2+y^2}\right)}^2 \\ \Rightarrow {\left(x-8\right)}^2={\left(x-8\right)}^2+y^2 \\ \Rightarrow 0=y^2 \\ \Rightarrow 0=y \end{gathered}$$

So the complex number $z$ lies on the equation $y=0$ which is an equation of straight line and represents the real axis in the complex plane.

So the point $z$ will lie on a straight line.

Hence, the correct option is C.