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

$3/4$ .