Given the equation of two circles `x^(2)+y^(2)=r^(2) and x^(2) +y^(2) -10x +16=0`. the value of r such that they intersect in real and distinct points is given by

Options

A. `2 lt r lt 8`
B. `r =2 or r = 8`
C. `r lt 2 or r gt 8`
D. none of these

Answer

To solve the problem, we need to determine the value of \( r \) such that the two circles intersect at real and distinct points.

### Step-by-Step Solution:

1. **Identify the equations of the circles:**
- The first circle is given by the equation:
\[ x^2 + y^2 = r^2 \]
This circle has its center at \( (0, 0) \) and radius \( r \).
- The second circle is given by the equation:
\[ x^2 + y^2 - 10x + 16 = 0 \]
We can rewrite this equation in standard form by completing the square.

2. **Rewrite the second circle's equation:**
- Rearranging the second circle's equation:
\[ x^2 - 10x + y^2 + 16 = 0 \]
- Completing the square for the \( x \) terms:
\[ (x^2 - 10x + 25) + y^2 = 25 - 16 \]
\[ (x - 5)^2 + y^2 = 9 \]
- This shows that the center of the second circle is \( (5, 0) \) and the radius \( r_2 = 3 \).

3. **Determine the distance between the centers:**
- The distance \( d \) between the centers of the two circles \( C_1(0, 0) \) and \( C_2(5, 0) \) is:
\[ d = \sqrt{(5 - 0)^2 + (0 - 0)^2} = 5 \]

4. **Set up the conditions for intersection:**
- For the circles to intersect at real and distinct points, the following condition must hold:
\[ |r_1 - r_2| < d < r_1 + r_2 \]
- Here, \( r_1 = r \) and \( r_2 = 3 \), so we can substitute these values into the inequalities:
\[ |r - 3| < 5 < r + 3 \]

5. **Solve the inequalities:**
- **First part:** \( |r - 3| < 5 \)
- This gives us two inequalities:
\[ -5 < r - 3 < 5 \]
- Adding 3 to all parts:
\[ -2 < r < 8 \]

- **Second part:** \( 5 < r + 3 \)
- Subtracting 3 from both sides:
\[ 2 < r \]

6. **Combine the inequalities:**
- From the two parts, we have:
\[ 2 < r < 8 \]

### Final Answer:
The value of \( r \) such that the two circles intersect at real and distinct points is:
\[ 2 < r < 8 \]