If `|(z-25)/(z-1)| = 5 ` , the value of `|z|` is -

Options

A. 3
B. 4
C. 5
D. 6

Explanation

Let `z = x + I y `
`:.| ( z- 25)/( z-1)| =5 rArr | ((x -25) + iy)/( ( x-1)+ i y ) | = 5 rArr | (x-25) + iy | = 5 | ( x - 1) + iy | `
`rArr sqrt((x=25)^(2) +y^(2)) = 5 sqrt((x-1)^(2) + y^(2))`
On square both sides, we get
`( x- 25)^(2) +y^(2) = 25 { ( x- 1)^(2) + y^(2) } rArr x^(2)- 50 + 625 + y^(2) = 25 x^(2) -50x +25 +25y^(2)`
`rArr 24 x^(2) + 24y^(2) = 600 rArr x^(2) +y^(2) = 25 `
`rArr sqrt(x^(2) + y^(2))=5 ` `[ Q | z| = sqrt((x^(2) + y^(2)))] rArr |z| =5 `