Question

Let $\alpha ,\beta \in R$ and $(1-\cos \alpha {)}^2+{\sin }^2\beta \ne 0$. Suppose $S=\{z\in C:z=\frac{1}{(1-\cos \alpha )+ik\sin \beta },k\in R-\{0\left\}\right\},$ where $i=\sqrt{-1}.$ If $z=x+iy,z\in S$ and $(x,y)$ lies on a circle, then

• A. the minimum radius of the circle is $\frac{1}{4}$ and corresponding centre is $(\frac{1}{4},0)$
• B. the minimum radius of the circle is $\frac{1}{2}$ and corresponding centre is $(\frac{1}{2},0)$
• C. $\alpha \ne 2n\pi$ and $\beta \in R$
• D. $\alpha =2n\pi$ and $\beta \ne n\pi$

Answer 1

Answer: (A), (C)

The correct options are
A the minimum radius of the circle is $\frac{1}{4}$ and corresponding centre is $(\frac{1}{4},0)$
C $\alpha \ne 2n\pi$ and $\beta \in R$

$x+iy=\frac{1}{(1-\cos \alpha )+ik\sin \beta }=\frac{(1-\cos \alpha )-ik\sin \beta }{(1-\cos \alpha {)}^2+k^2{\sin }^2\beta }\cdots \left(1\right)$

Equating the real and imaginary parts of the two sides, we get $x=\frac{(1-\cos \alpha )}{(1-\cos \alpha {)}^2+k^2{\sin }^2\beta }\cdots \left(2\right)$
$y=\frac{-k\sin \beta }{(1-\cos \alpha {)}^2+k^2{\sin }^2\beta }\cdots \left(3\right)$

We eliminate $k$ in these two equations.
Dividing $\left(3\right)$ by $\left(2\right)$, we have
$\frac{y}{x}=\frac{-k\sin \beta }{(1-\cos \alpha )}\cdots \left(4\right)$
which gives $k=\frac{-y(1-\cos \alpha )}{x\sin \beta }$
We can put this either in $\left(2\right)$ or $\left(3\right)$, which gives $x=\frac{(1-\cos \alpha )}{(1-\cos \alpha {)}^2+\frac{y^2(1-\cos \alpha {)}^2\cdot {\sin }^2\beta }{x^2{\sin }^2\beta }}$
$\Rightarrow x=\frac{x^2(1-\cos \alpha )}{(1-\cos \alpha {)}^2(x^2+y^2)}$
$\Rightarrow 1=\frac{x}{(1-\cos \alpha )(x^2+y^2)}$
$\Rightarrow x^2+y^2=\frac{x}{(1-\cos \alpha )}$
$\Rightarrow x^2-\frac{x}{a}+\frac{1}{4a^2}-\frac{1}{4a^2}+y^2=0(\text{Let}a=1-\cos \alpha )$
$\Rightarrow {(x-\frac{1}{2a})}^2+y^2={\left(\frac{1}{2a}\right)}^2$
The above equation represents the circle of centre $(\frac{1}{2a},0)$ and radius $\frac{1}{2a}$ when $\cos \alpha \ne 1$ i.e., $\alpha \ne 2n\pi$

$\Rightarrow r=\frac{1}{2a}=\frac{1}{2(1-\cos \alpha )}$
$\Rightarrow r_{min}=\frac{1}{4}$ when $\cos \alpha =-1$
This is a circle with minimum radius $=\frac{1}{4}$ and corresponding centre will be $(\frac{1}{4},0)$.