If one of the vertices of the square circumscribing the circle -z - 1- - -2 is 2 - -3i- find the order of vertices of the square-

The correct option is A z_2 = (1 √(3)) + i, z_3 = i√(3), z_4 = (√(3) + 1) i Here center is (1,0) is also the mid point of diagonal of square. ...

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Standard XII

Mathematics

Equation of Circle with (h,k) as Center

If one of the...

Question

If one of the vertices of the square circumscribing the circle $| z - 1 | = \sqrt{2}$ is $2 + \sqrt{3} i$ , find the order of vertices of the square.

z_{2} = (1 - \sqrt{3}) + i, z_{3} = - i \sqrt{3}, z_{4} = (\sqrt{3} + 1) - i

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z_{2} = (1 - \sqrt{3}) + i, z_{3} = - i \sqrt{3}, z_{4} = (\sqrt{3} + 1) + i

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z_{2} = (1 - \sqrt{3}) + i, z_{3} = i \sqrt{3}, z_{4} = (\sqrt{3} + 1) - i

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z_{2} = (1 - \sqrt{3}) - i, z_{3} = - i \sqrt{3}, z_{4} = (\sqrt{3} + 1) + i

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Solution

The correct option is A $z_{2} = (1 - \sqrt{3}) + i, z_{3} = - i \sqrt{3}, z_{4} = (\sqrt{3} + 1) - i$
Here center is $(1, 0)$ is also the mid-point of diagonal of square.
$\Rightarrow \frac{z_{1} + z_{2}}{2} = z_{0}$
Where $z_{0} = 1 + 0 i$
$\Rightarrow z_{2} = - \sqrt{3} i$ and $\frac{z_{3} - 1}{z_{1} - 1} = e^{\pm \frac{i π}{2}}$
$\Rightarrow z_{3} = 1 + (1 + \sqrt{3} i) . (cos \frac{π}{2} \pm i sin \frac{π}{2}) (∵ z_{1} = 2 + \sqrt{3} i) = 1 \pm i (1 + \sqrt{3} i) = (1 \mp \sqrt{3}) \pm i$
$z_{3} = (1 - \sqrt{3}) + i$ and $z_{4} = (1 + \sqrt{3}) - i$

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If one of the vertices of the square circumscribing the circle |z−1|=√2 is 2+√3i, find the order of vertices of the square.

If one of the vertices of the square circumscribing the circle $| z - 1 | = \sqrt{2}$ is $2 + \sqrt{3} i$ , find the order of vertices of the square.