Two circles of radii 10 cm and 17 cm . Intersecting each other at two points and the distances between their centres is 21 cm . Find the length of the common chords.

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of Circle 1 (r1) = 10 cm
- Radius of Circle 2 (r2) = 17 cm
- Distance between centers (d) = 21 cm

2. **Set Up the Geometry:**
- Let the centers of the circles be C1 and C2.
- Let M be the midpoint of the common chord AB.
- The distance from C1 to M is denoted as x, and the distance from C2 to M is (d - x) = (21 - x).

3. **Use the Pythagorean Theorem:**
- For Circle 1:
\[ AC1^2 = AM^2 + MC1^2 \]
\[ 10^2 = AM^2 + x^2 \quad \text{(1)} \]
\[ 100 = AM^2 + x^2 \quad \text{(1)} \]

- For Circle 2:
\[ AC2^2 = AM^2 + MC2^2 \]
\[ 17^2 = AM^2 + (21 - x)^2 \quad \text{(2)} \]
\[ 289 = AM^2 + (21 - x)^2 \quad \text{(2)} \]

4. **Expand Equation (2):**
\[ 289 = AM^2 + (441 - 42x + x^2) \]
Rearranging gives:
\[ 289 = AM^2 + 441 - 42x + x^2 \]
\[ AM^2 = 289 - 441 + 42x - x^2 \]
\[ AM^2 = -152 + 42x - x^2 \quad \text{(3)} \]

5. **Set Equations (1) and (3) Equal:**
\[ 100 - x^2 = -152 + 42x - x^2 \]
Canceling \( -x^2 \) from both sides:
\[ 100 = -152 + 42x \]
\[ 42x = 100 + 152 \]
\[ 42x = 252 \]
\[ x = \frac{252}{42} = 6 \text{ cm} \]

6. **Find AM using x:**
- Substitute x back into Equation (1):
\[ 100 = AM^2 + 6^2 \]
\[ 100 = AM^2 + 36 \]
\[ AM^2 = 100 - 36 = 64 \]
\[ AM = \sqrt{64} = 8 \text{ cm} \]

7. **Calculate the Length of the Common Chord AB:**
\[ AB = 2 \times AM = 2 \times 8 = 16 \text{ cm} \]

### Final Answer:
The length of the common chord AB is **16 cm**.