Radius of the circle=?

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1 Answer
Mar 2, 2018

3/434.

Explanation:

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.