Two tangents are drawn from any point P on a given line `L=0` to a given circle `S=0`, if these tangents touch the circle at point A and B, then locus of the circumcentre the `DeltaPAB` will be

Options

A. A pair of straight lines
B. A circle touching the given line `L=0`
C. A circle touching the given circle `S=0`
D. A line parallel to given line `L=0`

Explanation

To solve the problem, we need to find the locus of the circumcenter of triangle \( PAB \), where \( P \) is a point on a given line \( L = 0 \), and \( A \) and \( B \) are points where the tangents from \( P \) touch a given circle \( S = 0 \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let the given line be \( L: Ax + By + C = 0 \).
- Let the equation of the circle be \( S: (x - h)^2 + (y - k)^2 = r^2 \).
- Point \( P \) lies on line \( L \).

2. **Finding Points of Tangency**:
- From point \( P \), two tangents are drawn to the circle, touching it at points \( A \) and \( B \).
- The tangents from a point to a circle are equal in length.

3. **Circumcenter of Triangle \( PAB \)**:
- The circumcenter \( O \) of triangle \( PAB \) is the point where the perpendicular bisectors of sides \( PA \), \( PB \), and \( AB \) intersect.
- The circumcenter is equidistant from points \( P \), \( A \), and \( B \).

4. **Using the Property of Tangents**:
- The angle between the line \( PA \) and the radius \( OA \) (where \( O \) is the center of the circle) is \( 90^\circ \).
- Therefore, the circumcenter \( O \) lies on the line that is perpendicular to \( PA \) at point \( A \) and also on the line perpendicular to \( PB \) at point \( B \).

5. **Finding the Locus**:
- As point \( P \) moves along line \( L \), the points \( A \) and \( B \) will also move, but they will always lie on the circle.
- The locus of the circumcenter \( O \) can be derived from the fact that it is the midpoint of the segment joining \( P \) and the center of the circle \( O \) when the triangle \( PAB \) is isosceles (which it is, due to the equal tangents).

6. **Equation of the Locus**:
- The locus of the circumcenter will be a line parallel to \( L \).
- Thus, the locus can be expressed as \( L' = Ax + By + D = 0 \), where \( D \) is a constant that depends on the radius and position of the circle.

### Conclusion:
The locus of the circumcenter of triangle \( PAB \) is a line parallel to the given line \( L = 0 \).