Question

If $z$ lies on the curve $|z-3-4i|=3$, then least value of $|z|$ is

• A. $2$
• B. $3$
• C. $8$
• D. None of these

Answer 1

The correct option is A $2$

Let $z=x+iy$
Hence, $|(x-3)+(y-4)i|=3$
Therefore, $(x-3{)}^2+(y-4{)}^2=9$
Hence $z$ lies on the above circle.
Let,
$x-3=3cos\theta$
$x=3+3costheta$
$y-4=3sin\theta$
$y=4+3sin\theta$
Hence ,$z=3(1+cos\theta )+i(4+3sin\theta )$
$|z{|}^2=9(1+cos\theta {)}^2+(4+3sin\theta {)}^2$ ...(i)
$=9[1+co{s}^2\theta +2cos\theta ]+[16+24sin\theta +9si{n}^2\theta ]$
$=25+9+18cos\theta +24sin\theta$
$=34+18cos\theta +24sin\theta$
Differentiating with respect to $\theta$ and equating with 0, we get
$\Rightarrow -18sin\theta +24cos\theta =0$
$\Rightarrow 3sin\theta =4cos\theta$
$\Rightarrow tan\theta =\frac{4}{3}$
Hence,
$\Rightarrow sin\theta =\frac{\pm 4}{5}$

$\Rightarrow cos\theta =\frac{\pm 3}{5}$
Now Considering
$\Rightarrow sin\theta =\frac{-4}{5}$

$\Rightarrow cos\theta =\frac{-3}{5}$

$\Rightarrow z=3(1-\frac{3}{5})+i(4-\frac{12}{5})$

$\Rightarrow \frac{6}{5}+i\frac{8}{5}$
Hence
$\Rightarrow |z{|}_{min}=\frac{10}{5}$
$=2$