Question

question 48 (I)
please help me in this question ​

Answer 1

Answer:

(x+9)(x+1)+y²=0

Step-by-step explanation:

$given \: \: z = x + iy \: \: and \: \: | \frac{z - 3}{z + 3} | = 2$

$= | \frac{x - 3 + iy}{x + 3 + iy} | = 2$

$= \frac{ |x - 3 + iy| }{ |x + 3 + iy| } = 2$

$\frac{ { |x - 3 + iy| }^{2} }{ |x + 3 + iy| ^{2} } = 4$

$\frac{ {x}^{2} + {y}^{2} - 6x + 9 }{ {x}^{2} + {y}^{2} + 6x + 9 } = 4$

${x}^{2} + {y}^{2} - 6x + 9 = 4 {x}^{2} + 4 {y}^{2} + 24x +36$

On solving the equation, we get,

$(x + 9)(x + 1) + {y}^{2} = 0$