Question

Find all circle which are orthogonal to $|z|=1$ and $|z-1|=4$

• A. $|z+7+bi|=\sqrt{(24+b^2)}$
• B. $|z+7-bi|=\sqrt{(24+b^2)}$
• C. $|z+7+bi|=\sqrt{(48+b^2)}$
• D. $|z+7-bi|=\sqrt{(48+b^2)}$

Answer 1

The correct option is D $|z+7-bi|=\sqrt{(48+b^2)}$

Let $|z-\alpha |=k$ ........ (1)
(where $\alpha =a+bi$ and a, b, k $\in$ R) be a circle which cuts the circles
$|z|=1$ .......... (2)
and $|z-1|=4$ ........... (3)
orthogonally.
Then using the property that the sum of squares of their radii is equal to square of distance between centres. Thus the circle (1) will cut the circles (2) and (3) orthogonally if
$k^2+1=|\alpha -0{|}^2=\alpha \overline{\alpha }$
and $k^2+16=|\alpha -1{|}^2=(\alpha -1)(\overline{\alpha }-1)=\alpha \overline{\alpha }-(\alpha +\overline{\alpha })+1$
$\therefore 1-(\alpha +\overline{\alpha })-15=0$
or $\alpha +\overline{\alpha }=-14$
$\therefore 2a=-14\Rightarrow a=-7$
$\Rightarrow \alpha =a+ib$
$=-7+ib$
Also $k^2=|\alpha {|}^2-1=(-7{)}^2+b^2-1=b^2+48$
$\Rightarrow k=\sqrt{(b^2+48)}$
$\therefore$ Required family of circles is given by
$|z+7-bi|=\sqrt{(48+b^2)}$
Ans: D