Let the lines 2 - i z = 2 + i z ¯ and 2 + i z + i - 2 z ¯ - 4 i = 0 , (here i 2 = - 1 ) be normal to a circle C . If the line i z + z ¯ + 1 + i = 0 is tangent to this circle C , then its radius is :

Step-by-step solution

( 2 - i ) z = ( 2 + i ) z ¯ ⇒ ( 2 - i ) ( x + i y ) = ( 2 + i ) ( x - i y ) ⇒ 2 x - i x + 2 i y + y = 2 x + i x - 2 i y + y ⇒ 2 i x - 4 i y = 0 L 1 : x - 2 y = 0 ⇒ ( 2 + i ) z + ( i - 2 ) z ¯ - 4 i = 0 ⇒ ( 2 + i ) ( x + i y ) + ( i - 2 ) ( x - i y ) - 4 i = 0 . ⇒ 2 x + i x + 2 i y - y + i x - 2 x + y + 2 i y - 4 i = 0 ⇒ 2 i x + 4 i y - 4 i = 0 Solve L 1 and L 2 ,   4 y = 2 ,   y = 1 2 ∴   x = 1 Centre 1 , 1 2 L 3 : i z + z ¯ + 1 + i = 0 &