The operating characteristic of a reactance relay is given by X ≤ 1Ω, where X is the reactance calculated by the relay. Its operating characteristic in the ad mittance plane (G-B plane, where G and B denote conductance and susceptance, respectively, expressed in ) is given by:

The correct option is G² + (B + 0.5)² ≥ 1/4.

To understand why this is correct, we need to map the operating characteristic of the reactance relay from the impedance plane (R-X plane) to the admittance plane (G-B plane).
The operating characteristic of the reactance relay is given in the impedance plane by:
$X\le 1$
where X is the reactance of the line calculated by the relay.

Let the impedance be represented as:
$Z=R+jX$
The admittance Y is the reciprocal of impedance Z:
$Y=\frac{1}{Z}=G+jB$
where G represents the conductance and B represents the susceptance.

Expressing impedance Z in terms of admittance Y, we get:
$Z=\frac{1}{Y}=\frac{1}{G+jB}$
Multiplying the numerator and denominator by the complex conjugate of the denominator, (G - jB):
$Z=\frac{G-jB}{G^2+B^2}=\frac{G}{G^2+B^2}-j\frac{B}{G^2+B^2}$
Comparing the real and imaginary parts of Z = R + jX, we obtain the expression for reactance X:
$X=-\frac{B}{G^2+B^2}$

Substitute this expression for X into the inequality representing the relay's operating characteristic, X ≤ 1:
$-\frac{B}{G^2+B^2}\le 1$
Since G² + B² is always positive (for any non-zero admittance), we can multiply both sides of the inequality by G² + B² without changing the direction of the inequality sign:
$-B\le G^2+B^2$
Rearranging the terms:
$G^2+B^2+B\ge 0$

To write this in standard circle equation format, complete the square for the B terms by adding (1/2)² = 1/4 to both sides:
$G^2+\left(B^2+B+\frac{1}{4}\right)\ge \frac{1}{4}$
This simplifies directly to:
$G^2+{\left(B+0.5\right)}^2\ge \frac{1}{4}$
This equation represents the region outside (and including the boundary of) a circle in the G-B plane centered at (0, -0.5) with a radius of 0.5.