The equation $| 2 z + 3 | = | 3 z + 4 |$ on the complex plane where $z = x + i y, x, y \in R$

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represents a circle with centre $(\frac{6}{5}, 0)$ and radius is $\frac{2}{5}$

represents a circle with centre $(\frac{- 6}{5}, 0)$ and radius $\frac{1}{5}$

is such that ${| z |}_{\max} = \frac{7}{5}$

is such that ${| z |}_{\min} = 1$

answer is B, C, D.

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Detailed Solution

${(2 x + 3)}^{2} + 4 y^{2} = {(3 x + 4)}^{2} + 9 y^{2} \Rightarrow x^{2} + y^{2} + \frac{12}{5} x + \frac{7}{5} = 0$
Centre $= (- \frac{6}{5}, 0), r = \frac{1}{5}$
$\begin{array}{l} {| z |}_{\min} = O C - r = \frac{6}{5} - \frac{1}{5} = 1 \\ {| z |}_{\max} = O C + r = \frac{6}{5} + \frac{1}{5} = \frac{7}{5} \end{array}$