Let R denote the set of all real numbers. Let | vedclass

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(B) Given $|x + iy - (1 + 2i)| = 2|x + iy - 3i|$.Squaring both sides,we get $(x - 1)^2 + (y - 2)^2 = 4(x^2 + (y - 3)^2)$.Expanding the terms: $x^2 - 2

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$A, D$

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(B) Given $|x + iy - (1 + 2i)| = 2|x + iy - 3i|$.Squaring both sides,we get $(x - 1)^2 + (y - 2)^2 = 4(x^2 + (y - 3)^2)$.Expanding the terms: $x^2 - 2x + 1 + y^2 - 4y + 4 = 4(x^2 + y^2 - 6y + 9)$.$x^2 + y^2 - 2x - 4y + 5 = 4x^2 + 4y^2 - 24y + 36$.Rearranging terms: $3x^2 + 3y^2 + 2x - 20y + 31 = 0$.Dividing by $3$: $x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0$.The centre is $\left(-\frac{1}{3}, \frac{10}{3}\right)$.The radius is $\sqrt{\left(-\frac{1}{3}\right)^2 + \left(\frac{10}{3}\right)^2 - \frac{31}{3}} = \sqrt{\frac{1}{9} + \frac{100}{9} - \frac{93}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.Thus,statements $A$ and $D$ are true.

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