Two spheres of radii `r_(1) and r_(2)`, cut orthogonally. The radius of the common chord is 1`(r_(1)r_(2))/(sqrt""(r_(1)^(2)+r_(2)^(2)))`

Answer

To solve the problem of finding the radius of the common chord between two orthogonally intersecting spheres of radii \( r_1 \) and \( r_2 \), we will follow a systematic approach:

### Step-by-Step Solution

1. **Understanding the Geometry**:
- We have two spheres with centers \( C_1 \) and \( C_2 \) and radii \( r_1 \) and \( r_2 \) respectively.
- The spheres intersect orthogonally, meaning the angle between their radii at the intersection points is \( 90^\circ \).

**Hint**: Visualize the two spheres and their centers. The intersection forms a common chord.

2. **Identifying the Common Chord**:
- Let the points where the spheres intersect be \( A \) and \( B \). The line segment \( AB \) is the common chord.
- Let \( L \) be the midpoint of the chord \( AB \). The length \( AL \) is half the length of the chord.

**Hint**: Label the points clearly to keep track of the geometry.

3. **Using Trigonometric Relationships**:
- In triangle \( C_1AL \):
- \( AL = r_1 \sin(\theta) \) where \( \theta \) is the angle at \( C_1 \).
- In triangle \( C_2AL \):
- \( AL = r_2 \cos(\theta) \) since \( \angle C_2AL = 90^\circ - \theta \).

**Hint**: Remember the sine and cosine definitions in right triangles.

4. **Setting Up the Equations**:
- From the above relationships, we can write:
\[ AL = r_1 \sin(\theta) \quad \text{(1)} \]
\[ AL = r_2 \cos(\theta) \quad \text{(2)} \]

**Hint**: You can express \( \sin(\theta) \) and \( \cos(\theta) \) in terms of \( AL \).

5. **Using the Pythagorean Identity**:
- From trigonometric identities, we know:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
- Substituting equations (1) and (2) into this identity gives:
\[ \left(\frac{AL}{r_1}\right)^2 + \left(\frac{AL}{r_2}\right)^2 = 1 \]

**Hint**: This is a crucial step that connects the lengths with the radii.

6. **Simplifying the Equation**:
- Rearranging gives:
\[ \frac{AL^2}{r_1^2} + \frac{AL^2}{r_2^2} = 1 \]
- Factoring out \( AL^2 \):
\[ AL^2 \left(\frac{1}{r_1^2} + \frac{1}{r_2^2}\right) = 1 \]

**Hint**: Factor out common terms to simplify the equation.

7. **Solving for \( AL^2 \)**:
- Thus, we can express \( AL^2 \) as:
\[ AL^2 = \frac{1}{\frac{1}{r_1^2} + \frac{1}{r_2^2}} = \frac{r_1^2 r_2^2}{r_1^2 + r_2^2} \]

**Hint**: This step involves taking the reciprocal of the sum of fractions.

8. **Finding \( AL \)**:
- Taking the square root gives:
\[ AL = \frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}} \]

**Hint**: Ensure to take the positive root since lengths are positive.

9. **Finding the Radius of the Common Chord**:
- The length of the common chord \( AB \) is:
\[ AB = 2 \cdot AL = 2 \cdot \frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}} \]

**Hint**: Remember that the common chord length is twice the distance from the center to the chord.

### Final Result
The radius of the common chord is given by:
\[ \text{Radius of common chord} = \frac{r_1 r_2}{\sqrt{r_1^2 + r_2^2}} \]