Let z_1 = 6 + i and z_2 = 4 - 3i. Let z be a | vedclass

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(A) The condition $\arg \left( \frac{z - z_1}{z_2 - z} \right) = \frac{\pi}{2}$ implies that the angle subtended by the line segment joining $z_1$ and

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$|z - (5 - i)| = \sqrt{5}$

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(A) The condition $\arg \left( \frac{z - z_1}{z_2 - z} \right) = \frac{\pi}{2}$ implies that the angle subtended by the line segment joining $z_1$ and $z_2$ at $z$ is $\frac{\pi}{2}$.This means $z$ lies on a circle with the segment $z_1z_2$ as its diameter.The center $C$ of the circle is the midpoint of $z_1$ and $z_2$:$C = \frac{z_1 + z_2}{2} = \frac{(6 + i) + (4 - 3i)}{2} = \frac{10 - 2i}{2} = 5 - i$.The radius $r$ of the circle is half the distance between $z_1$ and $z_2$:$r = \frac{|z_1 - z_2|}{2} = \frac{|(6 + i) - (4 - 3i)|}{2} = \frac{|2 + 4i|}{2} = |1 + 2i| = \sqrt{1^2 + 2^2} = \sqrt{5}$.Thus,the equation of the circle is $|z - C| = r$,which is $|z - (5 - i)| = \sqrt{5}$.

Let $z_1 = 6 + i$ and $z_2 = 4 - 3i$. Let $z$ be a complex number such that $\arg \left( \frac{z - z_1}{z_2 - z} \right) = \frac{\pi}{2}$,then $z$ satisfies -

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