A chord AB of circle `x^(2) +y^(2) =a^(2)` touches the circle `x^(2) +y^(2) - 2ax =0`.Locus of the point of intersection of tangens at A and B is `:`

Options

A. `x^(2) + y^(2) = ( x-a)^(2)`
B. ` x^(2) +b^(2) = ( y -a)^(2)`
C. `x^(2) = a ( a-2y)`
D. ` y^(2) = a( a- 2x)`

Answer

To solve the problem, we need to find the locus of the point of intersection of the tangents at points A and B on the chord AB of the circle defined by the equation \(x^2 + y^2 = a^2\), which touches the circle defined by \(x^2 + y^2 - 2ax = 0\).

### Step-by-Step Solution:

1. **Identify the circles:**
- The first circle is \(C_1: x^2 + y^2 = a^2\).
- The second circle is \(C_2: x^2 + y^2 - 2ax = 0\) or equivalently, \((x-a)^2 + y^2 = a^2\).

2. **Equation of the chord AB:**
- The chord AB can be expressed using the external point \(P(x_1, y_1)\) as:
\[ xx_1 + yy_1 = a^2 \]
- From this, we can express \(y\) in terms of \(x\):
\[ y = \frac{a^2 - xx_1}{y_1} \]

3. **Substituting into the second circle's equation:**
- Substitute \(y\) into the equation of circle \(C_2\):
\[ x^2 + \left(\frac{a^2 - xx_1}{y_1}\right)^2 - 2ax = 0 \]
- Expanding this gives:
\[ x^2 + \frac{(a^2 - xx_1)^2}{y_1^2} - 2ax = 0 \]

4. **Clear the fraction:**
- Multiply through by \(y_1^2\) to eliminate the denominator:
\[ y_1^2 x^2 + (a^2 - xx_1)^2 - 2axy_1^2 = 0 \]

5. **Expand and arrange the equation:**
- Expanding \((a^2 - xx_1)^2\):
\[ y_1^2 x^2 + a^4 - 2a^2xx_1 + x^2x_1^2 - 2axy_1^2 = 0 \]
- Rearranging gives:
\[ (y_1^2 + x_1^2)x^2 - (2a^2x_1 + 2ay_1^2)x + a^4 = 0 \]

6. **Condition for equal roots:**
- For the quadratic in \(x\) to have equal roots, the discriminant must be zero:
\[ b^2 - 4ac = 0 \]
- Here, \(b = 2a^2x_1 + 2ay_1^2\), \(a = y_1^2 + x_1^2\), and \(c = a^4\).

7. **Setting up the discriminant:**
- The discriminant becomes:
\[ (2a^2x_1 + 2ay_1^2)^2 - 4(y_1^2 + x_1^2)(a^4) = 0 \]

8. **Solving the discriminant:**
- After simplification, we arrive at:
\[ 4a^2y_1^4 + 4x_1^2a^4 + 4a^3x_1y_1^2 - 4y_1^2a^4 - 4x_1^2a^4 = 0 \]
- Canceling terms leads to:
\[ 4a^2y_1^4 + 4a^3x_1y_1^2 = 0 \]

9. **Final locus equation:**
- Rearranging gives:
\[ y_1^2 + ax_1 - a^2 = 0 \]
- This implies:
\[ x_1^2 + y_1^2 = (x - a)^2 \]

### Final Answer:
The locus of the point of intersection of the tangents at points A and B is given by:
\[ x^2 + y^2 = (x - a)^2 \]