A variable line cuts x-axis at A, y-axis at B where ` OA = a, OB =b ` (O as origin) such that then the locus of circumcentre of `DeltaOAB` is-

Answer

To find the locus of the circumcenter of triangle OAB, where the line cuts the x-axis at point A and the y-axis at point B, we can follow these steps:

### Step 1: Identify the coordinates of points A and B
- The point A, where the line intersects the x-axis, has coordinates (a, 0).
- The point B, where the line intersects the y-axis, has coordinates (0, b).

### Step 2: Determine the circumcenter of triangle OAB
- Triangle OAB is a right triangle with the right angle at O (the origin).
- The circumcenter of a right triangle is located at the midpoint of the hypotenuse.

### Step 3: Find the coordinates of the midpoint of AB
- The coordinates of point A are (a, 0) and the coordinates of point B are (0, b).
- The midpoint P (circumcenter) of AB can be calculated using the midpoint formula:
\[ P\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = P\left( \frac{a + 0}{2}, \frac{0 + b}{2} \right) = P\left( \frac{a}{2}, \frac{b}{2} \right) \]
- Thus, the coordinates of the circumcenter P are \(\left( \frac{a}{2}, \frac{b}{2} \right)\).

### Step 4: Relate a and b using the given condition
- According to the problem, we have the relation \(a^2 + b^2 = 1\).

### Step 5: Substitute a and b in terms of h and k
- From the coordinates of P, we can express:
\[ a = 2h \quad \text{and} \quad b = 2k \]

### Step 6: Substitute these expressions into the equation
- Substitute \(a\) and \(b\) into the equation \(a^2 + b^2 = 1\):
\[ (2h)^2 + (2k)^2 = 1 \]
\[ 4h^2 + 4k^2 = 1 \]

### Step 7: Simplify the equation
- Divide the entire equation by 4:
\[ h^2 + k^2 = \frac{1}{4} \]

### Step 8: Write the locus in standard form
- If we replace \(h\) and \(k\) with \(x\) and \(y\), we get:
\[ x^2 + y^2 = \frac{1}{4} \]

### Conclusion
- The locus of the circumcenter of triangle OAB is a circle with center at the origin (0, 0) and radius \(\frac{1}{2}\).