Question

If the complex number z is to satisfy $|z|=3,|z-\left\{a\right(1+i)-i\}|\le 3$ and $|z+2a-(a+1)i|>3$ simultaneously for at least one z then find all a $\in$ R.

• A. $a\in (\frac{-1-\sqrt{71}}{2},\frac{-1-4\sqrt{11}}{5})\cup (\frac{-1+4\sqrt{11}}{5},\frac{-1+\sqrt{71}}{2})$
• B. $a\in (\frac{1-\sqrt{71}}{2},\frac{-1-4\sqrt{11}}{5})\cup (\frac{-1+4\sqrt{11}}{5},\frac{1+\sqrt{71}}{2})$
• C. $a\in (\frac{1-\sqrt{71}}{2},\frac{1-4\sqrt{11}}{5})\cup (\frac{1+4\sqrt{11}}{5},\frac{1+\sqrt{71}}{2})$
• D. $a\in (\frac{-1-\sqrt{71}}{2},\frac{1-4\sqrt{11}}{5})\cup (\frac{1+4\sqrt{11}}{5},\frac{-1+\sqrt{71}}{2})$

Answer 1

The correct option is B $a\in (\frac{1-\sqrt{71}}{2},\frac{-1-4\sqrt{11}}{5})\cup (\frac{-1+4\sqrt{11}}{5},\frac{1+\sqrt{71}}{2})$

All z at a time lie on a circle $|z|=3$, but inside and outside the circle $|z-\left\{a\right(1+i)-i|=3$ and $|z+2a-(a+1)i|=3$ respectively.
Let $z=x+iy$ then equation of circles are
$x^2+y^2=9$ ......... (1)
$(x-a{)}^2+(y-a+1{)}^2=9$ ........ (2)
and $(x+2a{)}^2+(y-a-1{)}^2=9$ ........ (3)
Circle (1) and (2) should cut or touch then distance between their centres $\le$ sum of their radii
$\Rightarrow \sqrt{(a-0{)}^2+(a-1-0{)}^2}\le 3+3$
$\Rightarrow a^2+(a-1{)}^2\le 36$
$\Rightarrow 2a^2-2a-35\le 0$
$\Rightarrow a^2-a-\frac{35}{2}\le 0$
$(a-\frac{1+\sqrt{71}}{2})(a-\frac{1-\sqrt{71}}{2})\le 0$
$\therefore \frac{1-\sqrt{71}}{2}\le a\le \frac{1+\sqrt{71}}{2}$ ..........(4)
Again circle (1) and (3) should not cut or touch then
Distance between their centres > sum fo their radii
$\sqrt{(-2a-0{)}^2+(a+1-0{)}^2}>3+3$
$\Rightarrow \sqrt{5a^2+2a+1}>6$
$\Rightarrow 5a^2+2a+1>36$
$\Rightarrow 5a^2+2a-35>0$
$\Rightarrow a^2+\frac{2a}{5}-7>0$
then $(a-\frac{-1-4\sqrt{11}}{5})(a-\frac{-1+4\sqrt{11}}{5})>0$
$\therefore a\in (-\infty ,\frac{-1-4\sqrt{11}}{5})\cup (\frac{-1+4\sqrt{11}}{5},\infty )$ ............ (5)
The common values of a satisfying (4) and (5) are
$a\in (\frac{1-\sqrt{71}}{2},\frac{-1-4\sqrt{11}}{5})\cup (\frac{-1+4\sqrt{11}}{5},\frac{1+\sqrt{71}}{2})$
Ans: B