Vikas TU
Last Activity: 8 Years ago
put z = x+ iy wwe get,
x^2 + y^2 = 4...........................(1)
The coordinate -1 + 5z becomes,
(5x – 1, 5y)
put it in eqn. (1) we get,
(5x – 1)^2 + (5y)^2 = 4
25x^2 + 1 – 10x + 25y^2 = 4
25(x^2 + y^2) – 10x – 3 = 0
x^2 + y^2 -2x/5 – 3/25 = 0
Hence the point lies on circle.