Let the lines (2−i)z=(2+i)zˉ and (2+i)z+(i−2)zˉ−4i=0 (here i2=−1) be normal to a circle C. If the line izˉ+z+1+i=0 is tangent to this circle C, then its radius is
Solutions · 1
Official solution
\[(2 - i)z = (2 + i)\overline z \]
\[ \Rightarrow (2 - i)(x + iy) = (2 + i)(x - iy)\]
\[ \Rightarrow 2x - ix + 2iy + y = 2x + ix - 2 - iy + y\]
\[ \Rightarrow 2ix - 4iy = 0\]
L1:x−2y=0
\[ \Rightarrow (2 + i)z + (i - 2)\overline z - 4i = 0\]
\[ \Rightarrow (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0\]
\[ \Rightarrow 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0\]
\[ \Rightarrow 2ix + 4iy - 4i = 0\]
L2:x+2y−2=0
Solve L1 and L2: x = 1, \[4y = 2,y = {1 \over 2}\]
\[ \therefore \] x=1
Centre \[\left( {1,{1 \over 2}} \right)\]
\[{L_3}:iz + \overline z + 1 + i = 0\]
\[ \Rightarrow i(x + iy) + x - iy + 1 + i = 0\]
\[ \Rightarrow ix - y + x - iy + 1 + i = 0\]
\[ \Rightarrow (x - y + 1) + i(x - y + 1) = 0\]
Radius = distance from \[\left( {1,{1 \over 2}} \right)\] to x−y+1=0
\[ \Rightarrow \] \[r = {{1 - {1 \over 2} + 1} \over {\sqrt 2 }}\]
\[ \Rightarrow \] \[r = {3 \over {2\sqrt 2 }}\]