Question

If the ratio (z-i)/(z-1) is purely imaginary, prove that the point z lies on the circle whose centre is the point (1/2)*(1+i) and radius is 1/√2​

Answer 1

Answer:

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Answer 2

Answer:

Equation of circle.

Center at ($\frac{1}{2} ,\frac{1}{2}$) and radius $\frac{1}{\sqrt{2} }$ Proved.

Step-by-step explanation:

Let us assume that z=x+iy.

We are given that (z-i)/(z-1) is purely imaginary.

Now, $\frac{z-i}{z-1}$

=$\frac{x+(y-1)i}{(x-1)+iy}$

{Rationalizing the denominator}

=$\frac{[x+(y-1)i][(x-1)-yi]}{(x-1)^{2}+y^{2}  }$

=$\frac{[x(x-1)+y(y-1)]+[(y-1)(x-1)-xy]i}{(x-1)^{2}+y^{2}  }$

=$\frac{[x(x-1)+y(y-1)]}{{(x-1)^{2}+y^{2}  }}+\frac{[(y-1)(x-1)-xy]}{{(x-1)^{2}+y^{2}  }}i$

Since, the above expression is purely imaginary, hence we can write

$\frac{[x(x-1)+y(y-1)]}{{(x-1)^{2}+y^{2}  }}$=0 {the real part of the expression will be zero}

⇒x(x-1)+y(y-1)=0

⇒x²-x+y²-y=0

⇒$(x-\frac{1}{2} )^{2} +(y-\frac{1}{2} )^{2}=\frac{1}{2}$

⇒$(x-\frac{1}{2} )^{2} +(y-\frac{1}{2} )^{2}=(\frac{1}{\sqrt{2} }) ^{2}$

Therefore the above equation is a circle with center at ($\frac{1}{2} ,\frac{1}{2}$) and radius $\frac{1}{\sqrt{2} }$. (Proved)