Question

Complex numbers which satisfy both the equations $|z-1-i|=\sqrt{2}$ and $|z+1+i|=2$ is/are

• A. $\left(\frac{1+\sqrt{7}}{8}\right)+i\left(\frac{1-\sqrt{7}}{8}\right)$
• B. $\left(\frac{1+\sqrt{7}}{4}\right)+i\left(\frac{1-\sqrt{7}}{4}\right)$
• C. $\left(\frac{1-\sqrt{7}}{8}\right)+i\left(\frac{1+\sqrt{7}}{8}\right)$
• D. $\left(\frac{1-\sqrt{7}}{4}\right)+i\left(\frac{1+\sqrt{7}}{4}\right)$

Answer 1

Answer: (B), (D)

The correct options are
B $\left(\frac{1+\sqrt{7}}{4}\right)+i\left(\frac{1-\sqrt{7}}{4}\right)$
D $\left(\frac{1-\sqrt{7}}{4}\right)+i\left(\frac{1+\sqrt{7}}{4}\right)$

The given equation represents circles
$(x-1{)}^2+(y-1{)}^2=2$
and $(x+1{)}^2+(y+1{)}^2=4$

$\Rightarrow C_1(1,1),r_1=\sqrt{2},C_2(-1,-1),r_2=2$
$C_1{C}_2=\sqrt{8}=2.8,r_1+r_2=2+\sqrt{2}=3.41$
$C_1{C}_2<r_1+r_2$ and also $C_1{C}_2>r_2-r_1$ and hence the two circles are intersecting at two points. The common two points will satisfy both.
Solving both equations,we get intersection points as $\left(\frac{1+\sqrt{7}}{4}\right)+i\left(\frac{1-\sqrt{7}}{4}\right)$ and $\left(\frac{1-\sqrt{7}}{4}\right)+i\left(\frac{1+\sqrt{7}}{4}\right)$