PA and PB are tangents to a circle S touching it at points A and B. C is a point on S in between A and B as shown in the figure. LCM is a tangent to S intersecting PA and PB in S at points L and M, respectively. Then the perimeter of the triangle PLM depends on o

Options

A. A,B,C and P
B. P but not on C
C. P and C only
D. the radius of S only

Explanation

To solve the problem, we need to find the perimeter of triangle PLM, where PA and PB are tangents to the circle S, and LCM is another tangent intersecting PA and PB at points L and M.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- PA and PB are tangents to the circle S at points A and B, respectively.
- C is a point on the circle S between A and B.
- LCM is a tangent to the circle S intersecting PA at L and PB at M.

2. **Identifying the Segments**:
- The lengths of the tangents from a point outside the circle to the points of tangency are equal. Thus, we have:
- \( PL = PA \) (since L is on PA)
- \( PM = PB \) (since M is on PB)

3. **Using the Tangent Properties**:
- Since LCM is a tangent to the circle, we can also say:
- \( PL + LC = LA \) (because LA is the entire tangent from P to A)
- \( CM + PM = BM \) (because BM is the entire tangent from P to B)

4. **Finding the Perimeter**:
- The perimeter of triangle PLM can be expressed as:
\[ \text{Perimeter} = PL + LM + PM \]
- From our earlier observations:
- \( PL = PA \)
- \( PM = PB \)
- Therefore, we can express LM in terms of the tangents:
\[ LM = LC + CM \]
- Substituting the values:
\[ \text{Perimeter} = PA + PB + (LC + CM) \]
- Since \( LC + CM \) can be expressed in terms of the tangents to the circle, we can simplify this further.

5. **Final Expression**:
- The perimeter of triangle PLM can be simplified to:
\[ \text{Perimeter} = PA + PB + (LA + BM) \]
- Since \( LA = PA \) and \( BM = PB \), we can conclude:
\[ \text{Perimeter} = PA + PB + PA + PB = 2(PA + PB) \]

### Conclusion:
Thus, the perimeter of triangle PLM depends on the lengths of the tangents PA and PB.