Two circles of radii 14 cm and 10 cm intersect each other, and the length of their common chord is 16 cm. What is the distance (in cm) between their centres?

Options

A. `6+4sqrt(33)`
B. `6+2sqrt(33)`
C. `6+5sqrt(33)`
D. `6+3sqrt(33)`

Answer

To solve the problem of finding the distance between the centers of two intersecting circles with given radii and the length of their common chord, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of Circle 1 (R1) = 14 cm
- Radius of Circle 2 (R2) = 10 cm
- Length of the common chord (AB) = 16 cm

2. **Calculate Half the Length of the Common Chord:**
- Let M be the midpoint of the common chord AB. Therefore, AM = MB = 16 cm / 2 = 8 cm.

3. **Use the Right Triangle Formed:**
- Draw the radii OA and OB from the centers O1 and O2 of the circles to the points A and B on the chord. The line segment OM (from the center of circle 1 to the midpoint M of the chord) is perpendicular to AB.
- In triangle O1AM, we can apply the Pythagorean theorem:
\[ O1A^2 = O1M^2 + AM^2 \]
- Here, O1A = R1 = 14 cm and AM = 8 cm.

4. **Substitute the Values:**
\[ 14^2 = O1M^2 + 8^2 \]
\[ 196 = O1M^2 + 64 \]
\[ O1M^2 = 196 - 64 = 132 \]
\[ O1M = \sqrt{132} = 2\sqrt{33} \text{ cm} \]

5. **Repeat for Circle 2:**
- Similarly, in triangle O2BM, we can use the Pythagorean theorem:
\[ O2B^2 = O2M^2 + MB^2 \]
- Here, O2B = R2 = 10 cm and MB = 8 cm.
\[ 10^2 = O2M^2 + 8^2 \]
\[ 100 = O2M^2 + 64 \]
\[ O2M^2 = 100 - 64 = 36 \]
\[ O2M = \sqrt{36} = 6 \text{ cm} \]

6. **Calculate the Distance Between the Centers:**
- The distance (d) between the centers O1 and O2 can be calculated as:
\[ d = O1M + O2M = 2\sqrt{33} + 6 \text{ cm} \]

### Final Answer:
The distance between the centers of the two circles is \( 6 + 2\sqrt{33} \) cm.