Let the lines (2-i)z = (2+i)bar{z} and (2+i)z | vedclass

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(D) The normal lines to a circle intersect at its center. Let $z = x + iy$. $(i)$ $(2-i)z = (2+i)\bar{z}$ $\Rightarrow (2-i)(x+iy) = (2+i)(x-iy)$ $\Ri

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$\frac{3}{2\sqrt{2}}$

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(D) The normal lines to a circle intersect at its center. Let $z = x + iy$. $(i)$ $(2-i)z = (2+i)\bar{z}$ $\Rightarrow (2-i)(x+iy) = (2+i)(x-iy)$ $\Rightarrow 2x + 2iy - ix + y = 2x - 2iy + ix + y$ $\Rightarrow 4iy - 2ix = 0$ $\Rightarrow y = \frac{x}{2}$. (ii) $(2+i)z + (i-2)\bar{z} - 4i = 0$ $\Rightarrow (2+i)(x+iy) + (i-2)(x-iy) - 4i = 0$ $\Rightarrow (2x + 2iy + ix - y) + (ix + y - 2x + 2iy) - 4i = 0$ $\Rightarrow 4iy + 2ix - 4i = 0$ $\Rightarrow x + 2y = 2$. Solving $(i)$ and (ii): Substitute $y = \frac{x}{2}$ into $x + 2y = 2$ $\Rightarrow x + x = 2$ $\Rightarrow x = 1, y = \frac{1}{2}$. So,the center of the circle is $(1, \frac{1}{2})$. (iii) The tangent line is $iz + \bar{z} + 1 + i = 0$ $\Rightarrow i(x+iy) + (x-iy) + 1 + i = 0$ $\Rightarrow ix - y + x - iy + 1 + i = 0$ $\Rightarrow (x-y+1) + i(x-y+1) = 0$ $\Rightarrow x - y + 1 = 0$. The radius $r$ is the perpendicular distance from the center $(1, \frac{1}{2})$ to the line $x - y + 1 = 0$: $r = \frac{|1 - \frac{1}{2} + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{3/2}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$.

Let the lines $(2-i)z = (2+i)\bar{z}$ and $(2+i)z + (i-2)\bar{z} - 4i = 0$ (where $i^2 = -1$) be normal to a circle $C$. If the line $iz + \bar{z} + 1 + i = 0$ is tangent to this circle $C$,then its radius is

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