O is the centre of a circle with radius 5 cm. LM is the diameter of the circle. P is a point on the plane of the circle such that LP = 6 cm and MP = 8 cm. Then P lies.

Options

A. on LM
B. outside the circle
C. inside the circle
D. on the circle

Answer

To solve the problem step by step, we will analyze the given information and apply the Pythagorean theorem.

### Step-by-Step Solution:

1. **Identify the Circle's Properties**:
- The center of the circle is O.
- The radius of the circle is 5 cm.
- Therefore, the diameter LM is twice the radius:
\[ LM = 2 \times 5 \text{ cm} = 10 \text{ cm} \]

2. **Position the Points**:
- Let L and M be the endpoints of the diameter, with O being the midpoint.
- We can place L at (-5, 0) and M at (5, 0) on a coordinate plane for simplicity. Thus, O is at (0, 0).

3. **Locate Point P**:
- We know that:
- \( LP = 6 \text{ cm} \)
- \( MP = 8 \text{ cm} \)

4. **Apply the Pythagorean Theorem**:
- According to the Pythagorean theorem, in triangle LPM:
\[ LP^2 + MP^2 = LM^2 \]
- Substitute the known lengths:
\[ 6^2 + 8^2 = LM^2 \]
- Calculate the squares:
\[ 36 + 64 = LM^2 \]
\[ 100 = LM^2 \]
- Since we have already calculated \( LM = 10 \text{ cm} \), we confirm:
\[ LM^2 = 10^2 = 100 \]

5. **Determine the Type of Triangle**:
- Since \( LP^2 + MP^2 = LM^2 \), triangle LPM is a right triangle with the right angle at point P.

6. **Conclusion**:
- According to the properties of circles, if a triangle is inscribed in a circle and one of its angles is a right angle, then the vertex of the right angle lies on the circumference of the circle.
- Therefore, point P lies on the circumference of the circle.

### Final Answer:
Point P lies on the circumference of the circle.

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