The two circles `x^2+ y^2=r^2` and `x^2+y^2-10x +16=0` intersect at two distinct points. Then

Options

A. `rlt2`
B. `rgt8`
C. `2ltrlt8`
D. `2lerle8`

Answer

To solve the problem of finding the range of \( r \) for the two circles to intersect at two distinct points, we will follow these steps:

### Step 1: Identify the equations of the circles
The first circle is given by:
\[ C_1: x^2 + y^2 = r^2 \]
The second circle can be rewritten from the given equation:
\[ C_2: x^2 + y^2 - 10x + 16 = 0 \implies x^2 + y^2 = 10x - 16 \]

### Step 2: Find the center and radius of the second circle
The equation of the second circle can be rearranged to:
\[ x^2 - 10x + y^2 = -16 \]
Completing the square for \( x \):
\[ (x - 5)^2 - 25 + y^2 = -16 \implies (x - 5)^2 + y^2 = 9 \]
From this, we can see that the center of the second circle \( C_2 \) is \( (5, 0) \) and the radius \( r_2 = 3 \).

### Step 3: Set the conditions for intersection
For the two circles to intersect at two distinct points, the distance between their centers must be less than the sum of their radii and greater than the absolute difference of their radii.

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

2. **Condition for intersection**:
- The sum of the radii: \( r + 3 > 5 \)
- The absolute difference of the radii: \( |r - 3| < 5 \)

### Step 4: Solve the inequalities
1. From \( r + 3 > 5 \):
\[ r > 2 \]

2. From \( |r - 3| < 5 \):
- This gives us two inequalities:
\[ r - 3 < 5 \implies r < 8 \]
\[ r - 3 > -5 \implies r > -2 \quad (\text{this is not restrictive since } r > 2) \]

### Step 5: Combine the results
From the inequalities, we have:
\[ 2 < r < 8 \]

### Conclusion
The range of \( r \) for which the two circles intersect at two distinct points is:
\[ r \in (2, 8) \]