Let `S_(1)` and `S_(2)` denote the circles `x^(2)+y^(2)+10x - 24y - 87 =0` and `x^(2) +y^(2)-10x -24y +153 = 0` respectively. The value of a for which the line `y = ax` contains the centre of a circle which touches `S_(2)` externally and `S_(1)` internally is

Options

A. `+-(3)/(10)`
B. `+-(1)/(5)`
C. `+-(sqrt(13))/(10)`
D. `+-(10)/(13)`

Explanation

To solve the problem, we need to analyze the two circles given by the equations \( S_1 \) and \( S_2 \), find their centers and radii, and then determine the value of \( a \) for the line \( y = ax \) that contains the center of a circle which touches \( S_2 \) externally and \( S_1 \) internally.

### Step 1: Find the centers and radii of the circles \( S_1 \) and \( S_2 \).

The equations of the circles are:
1. \( S_1: x^2 + y^2 + 10x - 24y - 87 = 0 \)
2. \( S_2: x^2 + y^2 - 10x - 24y + 153 = 0 \)

We can rewrite these equations in standard form by completing the square.

**For \( S_1 \):**
\[ x^2 + 10x + y^2 - 24y = 87 \]
Completing the square for \( x \):
\[ (x + 5)^2 - 25 \]
Completing the square for \( y \):
\[ (y - 12)^2 - 144 \]
Thus, we have:
\[ (x + 5)^2 + (y - 12)^2 = 256 \]
This gives us the center \( C_1(-5, 12) \) and radius \( R_1 = 16 \).

**For \( S_2 \):**
\[ x^2 - 10x + y^2 - 24y = -153 \]
Completing the square for \( x \):
\[ (x - 5)^2 - 25 \]
Completing the square for \( y \):
\[ (y - 12)^2 - 144 \]
Thus, we have:
\[ (x - 5)^2 + (y - 12)^2 = 16 \]
This gives us the center \( C_2(5, 12) \) and radius \( R_2 = 4 \).

### Step 2: Set up the relationship for the circle that touches \( S_2 \) externally and \( S_1 \) internally.

Let \( C \) be the center of the circle that touches \( S_2 \) externally and \( S_1 \) internally. The distances between the centers and the radii give us the following relationships:
1. \( C_1C + r = R_1 \) (touching \( S_1 \) internally)
2. \( C_2C - r = R_2 \) (touching \( S_2 \) externally)

Where \( r \) is the radius of the circle centered at \( C \).

### Step 3: Calculate the distances and set up equations.

The distance \( C_1C_2 \) is:
\[ C_1C_2 = \sqrt{(5 - (-5))^2 + (12 - 12)^2} = \sqrt{10^2} = 10 \]

From the relationships:
1. \( C_1C + r = 16 \)
2. \( C_2C - r = 4 \)

Let \( d = C_1C \) and \( d' = C_2C \). Then:
- From equation 1: \( d + r = 16 \) → \( r = 16 - d \)
- From equation 2: \( d' - r = 4 \) → \( r = d' - 4 \)

Setting these equal gives:
\[ 16 - d = d' - 4 \]
Thus:
\[ d' = d + 20 \]

### Step 4: Find the locus of point \( C \).

The locus of point \( C \) is an ellipse with foci at \( C_1 \) and \( C_2 \) and a major axis length of 20. The standard form of the ellipse centered at the midpoint of \( C_1 \) and \( C_2 \) is:
\[ \frac{(x - 0)^2}{100} + \frac{(y - 12)^2}{75} = 1 \]

### Step 5: Find the slope \( a \) for the line \( y = ax \).

The line \( y = ax \) is a tangent to the ellipse. The condition for tangency can be derived from the ellipse equation. The equation of the tangent line at point \( (x_0, y_0) \) on the ellipse is:
\[ \frac{x_0 x}{100} + \frac{(y_0 - 12)(y - 12)}{75} = 1 \]
Substituting \( y = ax \) into this equation and solving for \( a \) will yield the required value.

### Final Calculation:

After substituting and simplifying, we find that:
\[ m^2 = \frac{69}{100} \implies a = \pm \sqrt{\frac{13}{a}} \]

Thus, the value of \( a \) can be determined.