The circles `x^(2)+y^(2)-10x+16=0 and x^(2)+y^(2)=a^(2)` intersect at two distinct point if

Options

A. `a lt 2`
B. `2ltalt8`
C. `agt8`
D. `a=2`

Answer

To determine the condition for the circles \( x^2 + y^2 - 10x + 16 = 0 \) and \( x^2 + y^2 = a^2 \) to intersect at two distinct points, we will follow these steps:

### Step 1: Rewrite the equations of the circles
The first circle can be rewritten in standard form. We start with:
\[ x^2 + y^2 - 10x + 16 = 0 \]
To convert this into standard form, we complete the square for the \( x \) terms:
\[ (x^2 - 10x) + y^2 + 16 = 0 \]
Completing the square:
\[ (x - 5)^2 - 25 + y^2 + 16 = 0 \]
This simplifies to:
\[ (x - 5)^2 + y^2 = 9 \]
Thus, the center of the first circle is \( (5, 0) \) and the radius is \( 3 \).

The second circle is already in standard form:
\[ x^2 + y^2 = a^2 \]
This means its center is \( (0, 0) \) and its radius is \( a \).

### Step 2: Determine the distance between the centers
The distance \( d \) between the centers of the two circles is given by:
\[ d = \sqrt{(5 - 0)^2 + (0 - 0)^2} = \sqrt{25} = 5 \]

### Step 3: Apply the condition for intersection
For 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:
\[ |r_1 - r_2| < d < r_1 + r_2 \]
Where \( r_1 = 3 \) (radius of the first circle) and \( r_2 = a \) (radius of the second circle).

This gives us two inequalities:
1. \( |3 - a| < 5 \)
2. \( 5 < 3 + a \)

### Step 4: Solve the inequalities
**Inequality 1:**
\[ -5 < 3 - a < 5 \]
This can be split into two parts:
1. \( 3 - a > -5 \) which simplifies to \( a < 8 \)
2. \( 3 - a < 5 \) which simplifies to \( a > -2 \)

Since \( a \) is a radius, we only consider positive values:
\[ a < 8 \]

**Inequality 2:**
\[ 5 < 3 + a \]
This simplifies to:
\[ a > 2 \]

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

### Conclusion
The circles intersect at two distinct points if:
\[ a \in (2, 8) \]