Question:
The locus of a point \(z\) satisfying
\[
|z|^2=\operatorname{Re}(z)
\]
is a circle with centre
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
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The Correct Option is C
Solution and Explanation
Step 1: Represent the complex number in standard form.
Let \[ z=x+iy \] Then, \[ |z|^2=x^2+y^2 \] and \[ \operatorname{Re}(z)=x \] So the given equation becomes \[ x^2+y^2=x \]
Step 2: Rearrange the equation.
Bring all terms to one side: \[ x^2-x+y^2=0 \]
Step 3: Complete the square.
\[ x^2-x+\frac14+y^2=\frac14 \] Therefore, \[ \left(x-\frac12\right)^2+y^2=\left(\frac12\right)^2 \]
Step 4: Compare with the standard equation of a circle.
The standard form of a circle is \[ (x-a)^2+(y-b)^2=r^2 \] where the centre is \[ (a,b) \] Comparing, \[ \left(x-\frac12\right)^2+y^2=\left(\frac12\right)^2 \] we get \[ a=\frac12,\qquad b=0 \]
Step 5: Identify the centre.
Hence, the circle has centre \[ \left(\frac12,0\right) \]
Step 6: Final conclusion.
Therefore, the required centre is \[ \boxed{\left(\frac12,0\right)} \]
Let \[ z=x+iy \] Then, \[ |z|^2=x^2+y^2 \] and \[ \operatorname{Re}(z)=x \] So the given equation becomes \[ x^2+y^2=x \]
Step 2: Rearrange the equation.
Bring all terms to one side: \[ x^2-x+y^2=0 \]
Step 3: Complete the square.
\[ x^2-x+\frac14+y^2=\frac14 \] Therefore, \[ \left(x-\frac12\right)^2+y^2=\left(\frac12\right)^2 \]
Step 4: Compare with the standard equation of a circle.
The standard form of a circle is \[ (x-a)^2+(y-b)^2=r^2 \] where the centre is \[ (a,b) \] Comparing, \[ \left(x-\frac12\right)^2+y^2=\left(\frac12\right)^2 \] we get \[ a=\frac12,\qquad b=0 \]
Step 5: Identify the centre.
Hence, the circle has centre \[ \left(\frac12,0\right) \]
Step 6: Final conclusion.
Therefore, the required centre is \[ \boxed{\left(\frac12,0\right)} \]
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