If the two circles `x^(2) + y^(2) =4` and `x^(2) +y^(2) - 24x - 10y +a^(2) =0, a in I `, have exactly two common tangents then the number of possible integral values of a is `:`

A. 0
B. 2
C. 11
D. 13

Options

A. 0
B. 2
C. 11
D. 13

Answer

To solve the problem, we need to determine the number of integral values of \( a \) such that the two circles have exactly two common tangents.

### Step 1: Identify the equations of the circles
The first circle is given by:
\[ x^2 + y^2 = 4 \]
This can be rewritten in standard form, where the center \( C_1 \) is \( (0, 0) \) and the radius \( r_1 = 2 \).

The second circle is given by:
\[ x^2 + y^2 - 24x - 10y + a^2 = 0 \]
We can rearrange this into standard form:
\[ (x^2 - 24x) + (y^2 - 10y) + a^2 = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x - 12)^2 - 144 + (y - 5)^2 - 25 + a^2 = 0 \]
This simplifies to:
\[ (x - 12)^2 + (y - 5)^2 = 169 - a^2 \]
Thus, the center \( C_2 \) is \( (12, 5) \) and the radius \( r_2 = \sqrt{169 - a^2} \).

### Step 2: Calculate the distance between the centers
The distance \( d \) between the centers \( C_1 \) and \( C_2 \) is calculated as:
\[ d = \sqrt{(12 - 0)^2 + (5 - 0)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

### Step 3: Conditions for exactly two common tangents
For two circles to have exactly two common tangents, the following conditions must be satisfied:
1. The distance between the centers \( d \) must be greater than the difference of the radii:
\[ d > |r_1 - r_2| \]
2. The distance between the centers \( d \) must be less than the sum of the radii:
\[ d < r_1 + r_2 \]

Substituting the values we have:
- \( r_1 = 2 \)
- \( r_2 = \sqrt{169 - a^2} \)
- \( d = 13 \)

### Step 4: Set up the inequalities
From the first condition:
\[ 13 > |2 - \sqrt{169 - a^2}| \]
This leads to two cases:
1. \( 13 > 2 - \sqrt{169 - a^2} \)
2. \( 13 > \sqrt{169 - a^2} - 2 \)

From the first case:
\[ \sqrt{169 - a^2} > -11 \quad \text{(always true since radius is positive)} \]

From the second case:
\[ \sqrt{169 - a^2} < 15 \implies 169 - a^2 < 225 \implies a^2 > -56 \quad \text{(always true)} \]

Now, we focus on the second condition:
\[ 13 < 2 + \sqrt{169 - a^2} \]
This simplifies to:
\[ 11 < \sqrt{169 - a^2} \implies 121 < 169 - a^2 \implies a^2 < 48 \]

### Step 5: Combine the conditions
We now have:
1. \( a^2 < 48 \)
2. \( a^2 > 0 \) (since radius must be positive)

Thus, \( a \) must lie in the interval:
\[ -\sqrt{48} < a < \sqrt{48} \]
Calculating \( \sqrt{48} \approx 6.93 \), we find:
\[ -6 < a < 6 \]
The integral values of \( a \) in this range are:
\[ -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 \]
This gives us a total of 13 integral values.

### Final Answer
The number of possible integral values of \( a \) is:
\[ \boxed{13} \]