Let the lines (2−i)z=(2+i)z¯ and (2+i)z+(i−2)z¯=4i be a normal to circle ‘C’ . If the line iz+z¯+1+i=0 is a tangent to the circle ‘ C ‘ then its radius =

1 Normal’s are y=x2 and x+2y−2=0 are normals of circle∴ center=(1,12)Tangent is x-y+1=0∴r=|1−12+1|12+12=3/22=322

Let the lines (2−i)z=(2+i)z¯ and (2+i)z+(i−2)z¯=4i be a normal to circle ‘C’ . If the line iz+z¯+1+i=0 is a tangent to the circle ‘ C ‘ then its radius =

1 Normal’s are y=x2 and x+2y−2=0 are normals of circle∴ center=(1,12)Tangent is x-y+1=0∴r=|1−12+1|12+12=3/22=322

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Let the lines $(2 - i) z = (2 + i) \bar{z} a n d (2 + i) z + (i - 2) \bar{z} = 4 i$ be a normal to circle
‘C’ . If the line $i z + \bar{z} + 1 + i = 0$ is a tangent to the circle ‘ C ‘ then its
radius =

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

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

$3 \sqrt{2}$

$\frac{3}{2}$

answer is A.

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Detailed Solution

1 Normal’s are
$\begin{array}{l} y = \frac{x}{2} a n d x + 2 y - 2 = 0 are normals of circle \\ ∴ center = (1, \frac{1}{2}) \\ Tangent i s x - y + 1 = 0 \end{array}$
$∴ r = \frac{| 1 - \frac{1}{2} + 1 |}{\sqrt{1^{2} + 1^{2}}} = \frac{3 / 2}{\sqrt{2}} = \frac{3}{2 \sqrt{2}}$