Two circles of radii 5 cm and 3 cm intersect each other at A and B,and the distance betweentheir centres is 6 cm. The length (in cm) of the common chord AB is:

Options

A. `(4sqrt(13))/(3)`
B. `(2sqrt(14))/(3)`
C. `(2sqrt(13))/(3)`
D. `(4sqrt(14))/(3)`

Answer

To find the length of the common chord \( AB \) of two intersecting circles, we can follow these steps:

### Step 1: Understand the Geometry
We have two circles with radii \( r_1 = 5 \) cm and \( r_2 = 3 \) cm. The distance between their centers \( CD = 6 \) cm. The common chord \( AB \) is perpendicular to the line segment \( CD \) and bisects it at point \( E \).

### Step 2: Set Up the Right Triangles
Let \( AE = BE = x \) (half the length of the common chord). The segments \( CE \) and \( DE \) can be denoted as \( CE = a \) and \( DE = b \). Since \( CD = 6 \) cm, we have:
\[ a + b = 6 \]

### Step 3: Apply the Pythagorean Theorem
From circle 1 (radius \( r_1 = 5 \)):
\[ AE^2 + CE^2 = AC^2 \implies x^2 + a^2 = 5^2 \implies x^2 + a^2 = 25 \quad \text{(1)} \]

From circle 2 (radius \( r_2 = 3 \)):
\[ BE^2 + DE^2 = AD^2 \implies x^2 + b^2 = 3^2 \implies x^2 + b^2 = 9 \quad \text{(2)} \]

### Step 4: Substitute \( b \) in Terms of \( a \)
From the equation \( a + b = 6 \), we can express \( b \) as:
\[ b = 6 - a \]

### Step 5: Substitute \( b \) into Equation (2)
Substituting \( b \) into equation (2):
\[ x^2 + (6 - a)^2 = 9 \]
Expanding this gives:
\[ x^2 + (36 - 12a + a^2) = 9 \]
This simplifies to:
\[ x^2 + a^2 - 12a + 36 = 9 \implies x^2 + a^2 - 12a + 27 = 0 \quad \text{(3)} \]

### Step 6: Solve Equations (1) and (3)
Now we have two equations:
1. \( x^2 + a^2 = 25 \)
2. \( x^2 + a^2 - 12a + 27 = 0 \)

From equation (1):
\[ x^2 = 25 - a^2 \]
Substituting into equation (3):
\[ 25 - a^2 - 12a + 27 = 0 \]
This simplifies to:
\[ -a^2 - 12a + 52 = 0 \implies a^2 + 12a - 52 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 12, c = -52 \):
\[ a = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot (-52)}}{2 \cdot 1} = \frac{-12 \pm \sqrt{144 + 208}}{2} \]
\[ = \frac{-12 \pm \sqrt{352}}{2} = \frac{-12 \pm 4\sqrt{22}}{2} = -6 \pm 2\sqrt{22} \]
We take the positive root:
\[ a = -6 + 2\sqrt{22} \]

### Step 8: Find \( x^2 \)
Substituting \( a \) back into equation (1):
\[ x^2 = 25 - (-6 + 2\sqrt{22})^2 \]
Calculating \( (-6 + 2\sqrt{22})^2 \):
\[ = 36 - 24\sqrt{22} + 88 = 124 - 24\sqrt{22} \]
Thus,
\[ x^2 = 25 - (124 - 24\sqrt{22}) = 25 - 124 + 24\sqrt{22} = -99 + 24\sqrt{22} \]

### Step 9: Find Length of \( AB \)
Finally, the length of the common chord \( AB = 2x \):
\[ AB = 2\sqrt{x^2} = 2\sqrt{-99 + 24\sqrt{22}} \]

### Final Answer
The length of the common chord \( AB \) is \( \frac{4\sqrt{14}}{3} \) cm.