Two circles with same radius `r` intersect each other and one passes through the centre of the other. Then the length of the common chord is

Options

A. `r`
B. `sqrt(3)` r
C. `(sqrt(3) )/( 2)` r
D. `sqrt(5)` r

Answer

To find the length of the common chord of two intersecting circles with the same radius \( r \), where one circle passes through the center of the other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let the centers of the two circles be \( O \) and \( O' \).
- Since one circle passes through the center of the other, the distance \( OO' = r \).
- The radius of both circles is \( r \).

2. **Identifying Key Points**:
- Let \( A \) and \( B \) be the points where the two circles intersect.
- Let \( E \) be the midpoint of the common chord \( AB \).
- The line segment \( OE \) is perpendicular to \( AB \) and bisects it.

3. **Finding Lengths**:
- The distance \( OE \) can be determined using the right triangle \( OEC \) where \( C \) is the point where the line from \( O \) to \( O' \) meets the line \( AB \) perpendicularly.
- Since \( O' \) is at a distance \( r \) from \( O \), and \( E \) is the midpoint of \( AB \), we have \( OE = \frac{1}{2} OC \).

4. **Using Pythagorean Theorem**:
- In triangle \( OAE \) (where \( A \) is a point on the circumference of the circle), we can apply the Pythagorean theorem:
\[ OA^2 = OE^2 + EA^2 \]
- Here, \( OA = r \) (the radius), \( OE = \frac{1}{2} OC \) and \( OC = r \), thus \( OE = \frac{1}{2} r \).

5. **Calculating \( EA \)**:
- Substituting the values into the Pythagorean theorem:
\[ r^2 = \left(\frac{1}{2} r\right)^2 + EA^2 \]
- Simplifying gives:
\[ r^2 = \frac{1}{4} r^2 + EA^2 \]
- Rearranging leads to:
\[ EA^2 = r^2 - \frac{1}{4} r^2 = \frac{3}{4} r^2 \]
- Taking the square root gives:
\[ EA = \frac{\sqrt{3}}{2} r \]

6. **Finding Length of Common Chord \( AB \)**:
- Since \( AB = 2 \times EA \):
\[ AB = 2 \times \frac{\sqrt{3}}{2} r = \sqrt{3} r \]

### Final Answer:
The length of the common chord \( AB \) is \( \sqrt{3} r \).