Complex Numbers
Complex Numbers
star_batch_jee_advanced_2025
Grade None
Question:
Consider a square $OABC$ in the Argand plane, where 'O' is origin and $A=A\left(z_0\right)$. Then the equation of the circle that can be inscribed in this square is : (vertices of square are given in anti-clockwise order)
$|z-z_0\left(1+i\right)|=|z_0|$
$2\left|z-\frac{z_0\left(1+i\right)}{2}\right|=|z_0|$
$\left|z-\frac{z_0\left(1+i\right)}{2}\right|=|z_0|$
None of the above
Step-by-Step Solution
Key Concept: The center of an inscribed circle in a square is at the center of the square, and its radius is half the side length.
Place the square $OABC$ with $O$ at origin and $A$ at $z_0$. Since vertices are in anti-clockwise order: $O=0$, $A=z_0$, $B=z_0(1+i)$, $C=iz_0$. The side length is $|z_0|$. The center of the square is at the midpoint of diagonal $OB$, which is $\frac{z_0(1+i)}{2}$. The inradius of the square equals half the side length: $r=\frac{|z_0|}{2}$. Therefore, the inscribed circle has center $\frac{z_0(1+i)}{2}$ and radius $\frac{|z_0|}{2}$, giving $2\left|z-\frac{z_0(1+i)}{2}\right|=|z_0|$.
Correct Answer: 2