The length of the common chord of the two circles of radii 10, 24 whose centres are 26 units apart is

Options

A. `10/13` units
B. `240` units
C. `240/13` units
D. `120/13` units

Answer

To find the length of the common chord of two circles with radii 10 and 24 units, whose centers are 26 units apart, we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Circles and Identify Points**:
- Let the centers of the two circles be \( C_1 \) and \( C_2 \).
- The radius of circle \( C_1 \) is \( r_1 = 10 \) and the radius of circle \( C_2 \) is \( r_2 = 24 \).
- The distance between the centers \( C_1C_2 = d = 26 \).

2. **Label Points**:
- Let \( A \) and \( B \) be the points where the common chord intersects the circles.
- Let \( O \) be the midpoint of the common chord \( AB \).
- Let \( OA = H \) (the distance from the center of circle \( C_1 \) to the midpoint of the chord).
- Let \( OB = X \) (the distance from the center of circle \( C_2 \) to the midpoint of the chord).

3. **Apply Pythagorean Theorem**:
- In triangle \( C_1AO \):
\[ C_1A^2 = OA^2 + O C_1^2 \implies 10^2 = H^2 + X^2 \quad \text{(Equation 1)} \]
- In triangle \( C_2BO \):
\[ C_2B^2 = OB^2 + O C_2^2 \implies 24^2 = H^2 + (26 - X)^2 \quad \text{(Equation 2)} \]

4. **Substituting Values**:
- From Equation 1:
\[ 100 = H^2 + X^2 \quad \text{(1)} \]
- From Equation 2:
\[ 576 = H^2 + (26 - X)^2 \quad \text{(2)} \]

5. **Expand Equation 2**:
- Expanding Equation 2:
\[ 576 = H^2 + (676 - 52X + X^2) \]
- Rearranging gives:
\[ 576 = H^2 + 676 - 52X + X^2 \]
- Thus:
\[ 52X = 676 + H^2 - 576 + X^2 \]

6. **Subtract Equation 1 from Equation 2**:
- Subtracting Equation 1 from the expanded Equation 2:
\[ 576 - 100 = (H^2 + (26 - X)^2) - (H^2 + X^2) \]
- This simplifies to:
\[ 476 = 676 - 52X \]
- Rearranging gives:
\[ 52X = 200 \implies X = \frac{200}{52} = \frac{50}{13} \]

7. **Substituting Back to Find \( H \)**:
- Substitute \( X \) back into Equation 1:
\[ 100 = H^2 + \left(\frac{50}{13}\right)^2 \]
- Calculate \( \left(\frac{50}{13}\right)^2 = \frac{2500}{169} \):
\[ H^2 = 100 - \frac{2500}{169} = \frac{16900 - 2500}{169} = \frac{14400}{169} \]
- Thus:
\[ H = \frac{120}{13} \]

8. **Finding the Length of the Common Chord**:
- The length of the common chord \( AB \) is given by:
\[ AB = 2 \times OA = 2H = 2 \times \frac{120}{13} = \frac{240}{13} \]

### Final Answer:
The length of the common chord is \( \frac{240}{13} \).