A circle S=0 touches a circle `S_(1)=x^(2)+y^(2)-4x+6y-23=0` internally and the circle `S_(2)=x^(2)+y^(2)-4x+8y+19=0` externally. The locus of centre S=0 is a conic whose eccentricity is e and `r_(1) and r_(2)` be the radius of `S_(1)` & `S_(2) ` respectively and `[*]` denotes greatest integer, then

Options

A. `r_(1)+r_(2)=7`
B. `r_(1)-r_(2)=5`
C. `[(1)/(e)]=7`
D. none

Answer

To solve the problem, we need to find the locus of the center of the circle \( S = 0 \) that touches the circles \( S_1 \) and \( S_2 \) under the given conditions. Let's break down the solution step by step.

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

The equations of the circles are given as:
- \( S_1: x^2 + y^2 - 4x + 6y - 23 = 0 \)
- \( S_2: x^2 + y^2 - 4x + 8y + 19 = 0 \)

To find the center and radius of each circle, we can rewrite the equations in standard form.

**Circle \( S_1 \):**
1. Rearranging the equation:
\[ (x^2 - 4x) + (y^2 + 6y) = 23 \]
2. Completing the square:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 = 23 \]
\[ (x - 2)^2 + (y + 3)^2 = 36 \]
Thus, the center \( C_1 \) is \( (2, -3) \) and the radius \( R_1 = 6 \).

**Circle \( S_2 \):**
1. Rearranging the equation:
\[ (x^2 - 4x) + (y^2 + 8y) = -19 \]
2. Completing the square:
\[ (x - 2)^2 - 4 + (y + 4)^2 - 16 = -19 \]
\[ (x - 2)^2 + (y + 4)^2 = 1 \]
Thus, the center \( C_2 \) is \( (2, -4) \) and the radius \( R_2 = 1 \).

### Step 2: Set up the conditions for the locus of the center of circle \( S = 0 \).

Let \( C \) be the center of the circle \( S = 0 \) with coordinates \( (h, k) \). The conditions for touching the circles are:
1. \( C \) touches \( S_1 \) internally:
\[ \text{Distance}(C, C_1) = R_1 - r \]
where \( r \) is the radius of \( S = 0 \).

2. \( C \) touches \( S_2 \) externally:
\[ \text{Distance}(C, C_2) = R_2 + r \]

### Step 3: Write the distance equations.

From the centers and radii:
1. Distance to \( C_1 \):
\[ \sqrt{(h - 2)^2 + (k + 3)^2} = 6 - r \]
2. Distance to \( C_2 \):
\[ \sqrt{(h - 2)^2 + (k + 4)^2} = 1 + r \]

### Step 4: Square both equations to eliminate the square roots.

1. Squaring the first equation:
\[ (h - 2)^2 + (k + 3)^2 = (6 - r)^2 \]
2. Squaring the second equation:
\[ (h - 2)^2 + (k + 4)^2 = (1 + r)^2 \]

### Step 5: Expand and simplify both equations.

1. From the first equation:
\[ (h - 2)^2 + (k + 3)^2 = 36 - 12r + r^2 \]
2. From the second equation:
\[ (h - 2)^2 + (k + 4)^2 = 1 + 2r + r^2 \]

### Step 6: Set the two equations equal to each other.

Since both equal \( (h - 2)^2 \) we can set them equal:
\[ 36 - 12r + r^2 + 6k + 9 = 1 + 2r + r^2 + 8k + 16 \]
This simplifies to:
\[ 36 - 12r + 6k + 9 = 1 + 2r + 8k + 16 \]

### Step 7: Rearranging gives us the locus equation.

Rearranging gives:
\[ 20 = 14k + 14r \]
Thus, the locus of \( C \) is a conic section.

### Step 8: Determine the eccentricity \( e \).

The locus is an ellipse, and we can find the eccentricity using the relationship between the distances and the semi-major axis \( a \) and semi-minor axis \( b \).

1. The total distance between the two foci \( C_1 \) and \( C_2 \) is \( R_1 + R_2 = 7 \).
2. The semi-major axis \( a = \frac{7}{2} \).
3. The distance between the foci is \( d = \sqrt{(2 - 2)^2 + (-3 + 4)^2} = 1 \).
4. The distance between the foci \( 2c = 1 \) implies \( c = \frac{1}{2} \).

Using the relationship \( e = \frac{c}{a} \):
\[ e = \frac{1/2}{7/2} = \frac{1}{7} \]

### Final Result

The eccentricity \( e \) is \( \frac{1}{7} \).