Let $A$ and $B$ be variable points on $x-$axis and $y-$axis respectively such that the line segment $AB$ is in the first quadrant and of a fixed length $2d$ . Let $C$ be the mid-point of $AB$ and $P$ be a point such that

[b](a)[/b] $P$ and the origin are on the opposite sides of $AB$ and,

[b](b)[/b] $PC$ is a line segment of length $d$ which is perpendicular to $AB$ .

Find the locus of $P$ .
We are given that $A$ and $B$ are variable points on the $x$-axis and $y$-axis respectively, such that the line segment $AB$ is in the first quadrant and has a fixed length $2d$. Let $C$ be the midpoint of $AB$, and $P$ be a point such that:
1. $P$ and the origin are on opposite sides of $AB$.
2. $PC$ is a line segment of length $d$ which is perpendicular to $AB$.

We need to find the locus of $P$.

### Solution 1:

1. **Identify Coordinates of Points:**
   - Let $A = (a, 0)$ and $B = (0, b)$.
   - Since $AB$ has a fixed length $2d$, we have:
     \[
     \sqrt{a^2 + b^2} = 2d \implies a^2 + b^2 = 4d^2
     \]

2. **Find the Midpoint $C$:**
   - The coordinates of $C$ are:
     \[
     C = \left( \frac{a}{2}, \frac{b}{2} \right)
     \]

3. **Determine the Position of $P$:**
   - Since $PC$ is perpendicular to $AB$ and has length $d$, and $P$ and the origin are on opposite sides of $AB$, $P$ must lie on the line perpendicular to $AB$ passing through $C$.
   - The slope of $AB$ is $-\frac{b}{a}$, so the slope of the line perpendicular to $AB$ is $\frac{a}{b}$.
   - The equation of the line through $C$ with slope $\frac{a}{b}$ is:
     \[
     y - \frac{b}{2} = \frac{a}{b} \left( x - \frac{a}{2} \right)
     \]

4. **Calculate Coordinates of $P$:**
   - Since $PC = d$, we use the distance formula:
     \[
     \sqrt{\left( x - \frac{a}{2} \right)^2 + \left( y - \frac{b}{2} \right)^2} = d
     \]
   - Substituting $y$ from the line equation into the distance formula, we solve for $x$ and $y$.

5. **Conclude the Locus:**
   - By symmetry and the given conditions, $P$ must lie on the line $y = x$.
   - The distance $OP$ can only be in the interval $[d\sqrt{2}, 2d]$.

### Solution 2:

1. **Use Complex Numbers:**
   - Let $A = d \cos \theta$ and $B = 2id \sin \theta$.
   - Then $C = d e^{i\theta}$.

2. **Rotate $A$ about $C$:**
   - Since $\angle PCA = 90^\circ$, we rotate $A$ about $C$ anticlockwise by $90^\circ$ to get $P$.
   - This gives:
     \[
     P = (A - C)i + C = d(\sin \theta + \cos \theta)(i + 1) = d\sqrt{2}(i + 1)\sin\left(\theta + \frac{\pi}{4}\right)
     \]

3. **Determine the Locus:**
   - Clearly, this gives the locus of $P$ to be the line $y = x$.
   - Since $\theta \in \left[0, \frac{\pi}{2}\right]$, we have $|OP| \in [d\sqrt{2}, 2d]$.

\(\blacksquare\)

The final answer is \( \boxed{ y = x } \) with \( |OP| \in [d\sqrt{2}, 2d] \).