PR is a tangent at a point Q on a circle, in which Q is the centre and its radius is 8 cm. If OR = 16 cm and OP = 10 cm, then the length of PR is-

Options

A. 18 cm
B. 19 cm
C. 19.8 cm
D. 21.86 cm

Answer

To solve the problem, we need to find the length of the tangent PR to the circle at point Q, given the following information:

- The radius of the circle (OQ) = 8 cm
- The distance from the center to point R (OR) = 16 cm
- The distance from point O to point P (OP) = 10 cm

### Step-by-Step Solution:

1. **Identify the Right Triangle OPQ**:
- In triangle OPQ, O is the center of the circle, P is a point outside the circle, and Q is the point of tangency.
- OQ is the radius (8 cm), and OP is the distance from O to P (10 cm).

2. **Apply the Pythagorean Theorem in Triangle OPQ**:
- According to the Pythagorean theorem, we have:
\[ OP^2 = OQ^2 + PQ^2 \]
- Substituting the known values:
\[ 10^2 = 8^2 + PQ^2 \]
\[ 100 = 64 + PQ^2 \]
- Rearranging gives:
\[ PQ^2 = 100 - 64 = 36 \]
- Taking the square root:
\[ PQ = \sqrt{36} = 6 \text{ cm} \]

3. **Identify the Right Triangle OQR**:
- In triangle OQR, O is the center of the circle, Q is the point of tangency, and R is a point outside the circle.
- OR is the distance from O to R (16 cm).

4. **Apply the Pythagorean Theorem in Triangle OQR**:
- Again, using the Pythagorean theorem:
\[ OR^2 = OQ^2 + QR^2 \]
- Substituting the known values:
\[ 16^2 = 8^2 + QR^2 \]
\[ 256 = 64 + QR^2 \]
- Rearranging gives:
\[ QR^2 = 256 - 64 = 192 \]
- Taking the square root:
\[ QR = \sqrt{192} = 8\sqrt{3} \text{ cm} \]

5. **Calculate the Length of PR**:
- The length of the tangent PR can be found by adding the lengths of PQ and QR:
\[ PR = PQ + QR = 6 + 8\sqrt{3} \]

6. **Approximate the Value**:
- Using \( \sqrt{3} \approx 1.732 \):
\[ QR \approx 8 \times 1.732 = 13.856 \text{ cm} \]
- Therefore:
\[ PR \approx 6 + 13.856 = 19.856 \text{ cm} \]

### Final Answer:
The length of PR is approximately **19.856 cm**.