If |z - 2 - 3i| + |z + 2 - 6i| = 4,where i = | vedclass

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(B) The equation is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = 2 + 3i$ and $z_2 = -2 + 6i$.For this to represent an ellipse,the condition $

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$\phi$ (Empty set)

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(B) The equation is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = 2 + 3i$ and $z_2 = -2 + 6i$.For this to represent an ellipse,the condition $|z_1 - z_2| Here,$2a = 4$.Calculate the distance between the two fixed points:$|z_1 - z_2| = |(2 + 3i) - (-2 + 6i)| = |4 - 3i| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.Since the distance between the two points $(5)$ is greater than the sum of the distances $(4)$,the triangle inequality $|z - z_1| + |z - z_2| \ge |z_1 - z_2|$ is violated ($4 \ge 5$ is false).Therefore,there is no point $z$ that satisfies the given equation.Thus,the locus of $P(z)$ is $\phi$.

If $|z - 2 - 3i| + |z + 2 - 6i| = 4$,where $i = \sqrt{-1}$,then the locus of $P(z)$ is

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