Two circles whose radii are 10 cm and 8 cm,intersect each other and their common chord is 12 cm long. What is the distance between their centres ?

Options

A. 11.27 cm
B. 12.29 cm
C. 12.27 cm
D. 13.29 cm

Answer

To find the distance between the centers of two intersecting circles with given radii and a common chord length, 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) = 8 cm
- Length of the common chord (CD) = 12 cm

2. **Determine the Half-Length of the Common Chord:**
- Since the common chord is bisected by the line joining the centers of the circles, we can find the half-length of the chord:
\[ CE = \frac{CD}{2} = \frac{12}{2} = 6 \text{ cm} \]
where E is the midpoint of the chord CD.

3. **Set Up the Right Triangles:**
- Let A be the center of Circle 1 and B be the center of Circle 2. The line AB is perpendicular to the chord CD at point E.
- We can form two right triangles: Triangle ACE (for Circle 1) and Triangle BEC (for Circle 2).
- In Triangle ACE:
- AE = distance from center A to midpoint E
- CE = 6 cm (half of the chord)
- R1 = 10 cm (radius of Circle 1)

4. **Apply the Pythagorean Theorem in Triangle ACE:**
\[ AE^2 + CE^2 = R1^2 \]
\[ AE^2 + 6^2 = 10^2 \]
\[ AE^2 + 36 = 100 \]
\[ AE^2 = 100 - 36 = 64 \]
\[ AE = \sqrt{64} = 8 \text{ cm} \]

5. **Apply the Pythagorean Theorem in Triangle BEC:**
- In Triangle BEC:
- BE = distance from center B to midpoint E
- CE = 6 cm (half of the chord)
- R2 = 8 cm (radius of Circle 2)
\[ BE^2 + CE^2 = R2^2 \]
\[ BE^2 + 6^2 = 8^2 \]
\[ BE^2 + 36 = 64 \]
\[ BE^2 = 64 - 36 = 28 \]
\[ BE = \sqrt{28} = 2\sqrt{7} \approx 5.29 \text{ cm} \]

6. **Calculate the Distance Between the Centers (AB):**
- The total distance AB is the sum of AE and BE:
\[ AB = AE + BE = 8 + 5.29 = 13.29 \text{ cm} \]

### Final Answer:
The distance between the centers of the two circles is approximately **13.29 cm**.