Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the greater circle which is out side the inner circle is of length

Options

A. `2sqrt(2) cm `
B. `3sqrt(2) cm`
C. `2sqrt(3) cm`
D. `4sqrt(2) cm`

Answer

To solve the problem of finding the length of the biggest chord of the greater circle that is outside the inner circle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Circles and Their Centers**:
- Let the radius of the inner circle (smaller circle) be \( r_1 = 2 \) cm.
- Let the radius of the outer circle (greater circle) be \( r_2 = 3 \) cm.
- The centers of the circles will be denoted as \( O_1 \) for the inner circle and \( O_2 \) for the outer circle.

2. **Draw the Circles**:
- Draw the inner circle with radius 2 cm and the outer circle with radius 3 cm, ensuring that they touch each other internally.

3. **Determine the Distance Between the Centers**:
- Since the circles touch internally, the distance between the centers \( O_1 \) and \( O_2 \) is equal to the difference of their radii:
\[ O_1O_2 = r_2 - r_1 = 3 - 2 = 1 \text{ cm} \]

4. **Identify the Diameter of the Inner Circle**:
- The diameter of the inner circle is:
\[ \text{Diameter of inner circle} = 2 \times r_1 = 2 \times 2 = 4 \text{ cm} \]

5. **Find the Length of the Chord**:
- The biggest chord of the outer circle that is outside the inner circle is a tangent to the inner circle. This tangent will be perpendicular to the radius of the inner circle at the point of tangency.
- The distance from the center of the outer circle \( O_2 \) to the point of tangency on the inner circle is equal to the radius of the outer circle \( r_2 \) minus the radius of the inner circle \( r_1 \):
\[ O_2D = r_2 - r_1 = 3 - 2 = 1 \text{ cm} \]

6. **Use the Pythagorean Theorem**:
- In triangle \( O_2DB \) (where \( D \) is the point of tangency and \( B \) is the midpoint of the chord), we have:
\[ O_2D^2 + BD^2 = O_2B^2 \]
- Here, \( O_2D = 1 \) cm (as calculated), and \( O_2B = r_2 = 3 \) cm.
- Let \( BD = x \). Therefore:
\[ 1^2 + x^2 = 3^2 \]
\[ 1 + x^2 = 9 \]
\[ x^2 = 8 \implies x = \sqrt{8} = 2\sqrt{2} \]

7. **Calculate the Length of the Chord**:
- The total length of the chord \( BC \) is twice the length of \( BD \):
\[ BC = 2 \times BD = 2 \times 2\sqrt{2} = 4\sqrt{2} \text{ cm} \]

### Final Answer:
The length of the biggest chord of the greater circle which is outside the inner circle is \( 4\sqrt{2} \) cm.