Radius of the circle=?

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Answer 1 · 2018-03-02T13:05:11

# 3/4#.

Let, #z=x+iy; x,y in RR#.

#:. (iz-2)/(z-i)={i(x+iy)-2}/(x+iy-i)=(-y-2+ix)/{x+i(y-1)}#,

#=(-y-2+ix)/{x+i(y-1)}xx{x-i(y-1)}/{x-i(y-1)}#,

#={x(-y-2)+ix^2+i(y-1)(y+2)-i^2x(y-1)}/{x^2-i^2(y-1)^2}#,

#={x(-y-2+y-1)+i{x^2+(y-1)(y+2)}}/{x^2+(y-1)^2}#.

Clearly, #Im((iz-2)/(z-i))={x^2+(y-1)(y+2)}/{x^2+(y-1)^2}#,

#=-1............."[because, Given]"#,

#rArr {x^2+(y-1)(y+2)}+{x^2+(y-1)^2}=0#,

#rArr 2x^2+y^2+y-2+y^2-2y+1=0, or #,

# 2x^2+2y^2-y=1#.

#:. x^2+y^2-1/2y=1/2, or #,

# x^2+(y-1/4)^2=(1/4)^2+1/2=9/16=(3/4)^2#.

This represents a circle with centre at #(0,1/4)# and radius

#3/4#.