Question

The number of values of z which satisfies both the curves $|z-2-i|=\sqrt{5}$ &$|z+2+i|=5$ where $\sqrt{-1}=i$ is?

• A. $0$
• B. $1$
• C. $2$
• D. Infinitely many

Answer 1

The correct option is C $2$

From given, curves represents circles $|z-2-i|=\sqrt{5}\Rightarrow C_1(2+i),R_1=\sqrt{5}$
and $|z+2+i|=5\Rightarrow C_2(-2-i),R_2=5$
Now $\left|C_1{C}_2\right|=|2+i-(-2-i\left)\right|=|4+2i|=2\sqrt{5}....\left(i\right)$
Now $R_2-R_1=5-\sqrt{5}=\sqrt{5}(\sqrt{5}-1)<C_1{C}_2......\left(ii\right)$
$R_2+R_1=5+\sqrt{5}=\sqrt{5}(\sqrt{5}+1)<C_1{C}_2......\left(iii\right)$
Now $\sqrt{5}(\sqrt{5}-1)<\sqrt{5}\times 2<\sqrt{5}(\sqrt{5}+1)$ $\Rightarrow R_2-R_1<\left|C_1{C}_2\right|$
$<R_1+R_2\Rightarrow$
Circles intersect at two points