The circle `S_(1)(a_(1),b_(1)), r_(1)` touches externally the circles `S_(2) (a_(2), b_(2)), r_(2)`. If the tangent at their common point passes through origin, then

Options

A. `(a_(1)^(2)+a_(2)^(2)) +(b_(1)^(2)+b_(2)^(2))=r_(1)^(2)+r_(2)^(2)`
B. `(a_(1)^(2)-a_(2)^(2)) +(b_(1)^(2)-b_(2)^(2)) =r_(1)^(2)-r_(2)^(2)`
C. `(a_(1)^(2)-b_(1)^(2)) +(a_(2)^(2)+b_(2)^(2))=r_(1)^(2)+r_(2)^(2)`
D. `(a_(1)^(2)-b_(1)^(2))-(a_(2)^(2)+b_(2)^(2))=r_(1)^(2)-r_(2)^(2)`

Answer

To solve the problem, we need to analyze the given conditions about the circles and the tangents.

### Step-by-Step Solution:

1. **Understanding the Circles**:
We have two circles, \( S_1(a_1, b_1), r_1 \) and \( S_2(a_2, b_2), r_2 \). The first circle touches the second circle externally. This means the distance between their centers equals the sum of their radii:
\[ \sqrt{(a_2 - a_1)^2 + (b_2 - b_1)^2} = r_1 + r_2 \]

2. **Tangent Condition**:
The tangent at their common point passes through the origin. This implies that the line formed by the tangent can be expressed in the form \( y = mx \), where \( m \) is the slope.

3. **Using the Tangent Properties**:
For a circle centered at \( (a, b) \) with radius \( r \), the equation of the tangent line at point \( (x_0, y_0) \) on the circle is given by:
\[ (x - x_0)(x_0 - a) + (y - y_0)(y_0 - b) = r^2 \]
Since the tangent passes through the origin, we can substitute \( (x, y) = (0, 0) \) into the equation.

4. **Finding the Locus**:
By substituting \( (0, 0) \) into the tangent equation, we can derive a relationship involving \( a_1, b_1, r_1, a_2, b_2, r_2 \) and the slopes of the tangents.

5. **Using the Slope**:
The slopes of the tangents from point \( P(x_1, y_1) \) to the circles can be expressed using the angles \( \theta_1 \) and \( \theta_2 \). The relationship between the angles and the slopes gives us:
\[ \cot \theta_1 + \cot \theta_2 = c \]
which can be rewritten in terms of \( \tan \):
\[ \frac{1}{\tan \theta_1} + \frac{1}{\tan \theta_2} = c \]

6. **Setting Up the Quadratic**:
From the above relationships, we can derive a quadratic equation in terms of \( m \) (the slope of the tangent). The roots of this quadratic will represent the slopes of the tangents from point \( P \).

7. **Final Equation**:
After manipulating the expressions and substituting back, we arrive at the final equation representing the locus of point \( P \):
\[ C y^2 - a^2 = 2xy \]
This can be rearranged to form a standard equation of a conic section.

### Final Result:
The locus of point \( P \) is given by:
\[ C y^2 - 2xy - a^2 = 0 \]