Let z_1, z_2, z_3, omega, z_0, z'_0 be fixed | vedclass

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(A) The given condition $Arg\left( \frac{\omega - z_1}{z_2 - z_3} \right) = \frac{\pi}{2}$ implies that the vector $(\omega - z_1)$ is perpendicular t

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(A) The given condition $Arg\left( \frac{\omega - z_1}{z_2 - z_3} \right) = \frac{\pi}{2}$ implies that the vector $(\omega - z_1)$ is perpendicular to $(z_2 - z_3)$.Similarly,$(\omega - z_2) \perp (z_3 - z_1)$ and $(\omega - z_3) \perp (z_1 - z_2)$.This indicates that $\omega$ is the orthocentre of the triangle formed by $z_1, z_2, z_3$.Let $\Delta = \Delta z_1 z_2 z_3$. The circumcircle of $\Delta z_1 z_2 z_3$ has radius $R_1$ and center $z_0$.The circumcircle of $\Delta z_2 \omega z_3$ has radius $R_2$ and center $z'_0$.It is a known property of the orthocentre that the circumradius of the triangle formed by the orthocentre and two vertices is equal to the circumradius of the original triangle.Therefore,$R_1 = R_2$,which implies $\frac{R_1}{R_2} = 1$.

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