The points in the set {z in mathbb{C} : arg l | vedclass

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(A) Given that,$\arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}$.Let $z = x + iy$.The expression represents the locus of points $z$ such that the a

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(A) Given that,$\arg \left(\frac{z-2}{z-6i}ight) = \frac{\pi}{2}$.Let $z = x + iy$.The expression represents the locus of points $z$ such that the angle subtended by the segment joining $A(2, 0)$ and $B(0, 6)$ at $z$ is $\frac{\pi}{2}$.Using the property $\arg(z_1) - \arg(z_2) = \arg\left(\frac{z_1}{z_2}ight)$,we have:$\arg(z-2) - \arg(z-6i) = \frac{\pi}{2}$.Substituting $z = x + iy$:$	an^{-1}\left(\frac{y}{x-2}ight) - 	an^{-1}\left(\frac{y-6}{x}ight) = \frac{\pi}{2}$.Using the identity $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight) = \frac{\pi}{2}$,the denominator $1+AB$ must be $0$ as $A-B$ is finite.$1 + \left(\frac{y}{x-2}ight)\left(\frac{y-6}{x}ight) = 0$.$x(x-2) + y(y-6) = 0$.$x^2 - 2x + y^2 - 6y = 0$.This is the equation of a circle with center $(1, 3)$ and radius $\sqrt{1^2 + 3^2} = \sqrt{10}$.

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

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