Question:easy

The locus of a point \(z\) satisfying \[ |z|^2=\operatorname{Re}(z) \] is a circle with centre

Show Hint

For a complex number \[ z=x+iy, \] remember: \[ |z|^2=x^2+y^2 \] and \[ \operatorname{Re}(z)=x. \] After substitution, convert the equation into standard circle form by completing the square.
Updated On: Jun 22, 2026
  • \(\left(0,\frac12\right)\)
  • \(\left(-\frac12,0\right)\)
  • \(\left(\frac12,0\right)\)
  • \(\left(0,-\frac12\right)\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Write the condition in terms of $z = x + iy$.
We are given $|z|^2 = \text{Re}(z)$. Writing $z = x + iy$, we have $|z|^2 = x^2+y^2$ and $\text{Re}(z) = x$. So the condition becomes $x^2+y^2 = x$.
Step 2: Rearrange to standard circle form.
$x^2 - x + y^2 = 0$. Complete the square in $x$: \[\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + y^2 = 0 \implies \left(x-\frac{1}{2}\right)^2 + y^2 = \frac{1}{4}.\]
Step 3: Identify the circle.
This is a circle with center $\left(\dfrac{1}{2}, 0\right)$ and radius $\dfrac{1}{2}$.
Step 4: Locate the center on the Argand plane.
In the Argand (complex) plane, the center corresponds to the complex number $z_0 = \dfrac{1}{2} + 0 \cdot i = \dfrac{1}{2}$. In coordinate form, the center is $\left(\dfrac{1}{2}, 0\right)$.
Step 5: Verify a point on the locus.
At $z = 1$ (i.e., $x=1, y=0$): $|z|^2 = 1$ and $\text{Re}(z) = 1$. So $z=1$ lies on the locus. Check: $(1-\frac{1}{2})^2 + 0 = \frac{1}{4}$ ✓.
Step 6: State the answer.
The locus of $z$ is a circle with center at $\left(\dfrac{1}{2}, 0\right)$ and radius $\dfrac{1}{2}$. \[ \boxed{\text{Center} = \left(\frac{1}{2},\, 0\right)} \]
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