Statement-1: Tangents drawn from any point on the circle `x^(2)+y^(2)=25` to the ellipse `(x^(2))/(16)+(y^(2))/(9)=1` are at right angle Statement-2: The locus of the point of intersection of perpendicular tangents to an ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` is its director circle
`x^(2)+y^(2)=a^(2)+b^(2)`.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement -2 is not a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True

Explanation

To solve the given problem, we need to analyze the two statements provided and verify their correctness step by step.

### Step 1: Analyze Statement 1
**Statement 1:** Tangents drawn from any point on the circle \( x^2 + y^2 = 25 \) to the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \) are at right angles.

1. **Identify the Circle and Ellipse:**
- The circle is centered at the origin with a radius of 5 (since \( 25 = 5^2 \)).
- The ellipse has semi-major axis \( a = 4 \) (since \( \sqrt{16} = 4 \)) and semi-minor axis \( b = 3 \) (since \( \sqrt{9} = 3 \)).

2. **Find the Director Circle of the Ellipse:**
- The director circle of an ellipse is given by the equation \( x^2 + y^2 = a^2 + b^2 \).
- Here, \( a^2 + b^2 = 16 + 9 = 25 \).
- Therefore, the director circle for this ellipse is \( x^2 + y^2 = 25 \).

3. **Conclusion for Statement 1:**
- Since the circle \( x^2 + y^2 = 25 \) is the same as the director circle of the ellipse, it follows that the tangents drawn from any point on this circle to the ellipse will indeed be at right angles.

### Step 2: Analyze Statement 2
**Statement 2:** The locus of the point of intersection of perpendicular tangents to an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is its director circle \( x^2 + y^2 = a^2 + b^2 \).

1. **General Equation of Tangent to the Ellipse:**
- The equation of the tangent line to the ellipse in slope form is given by:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]

2. **Finding the Point of Intersection:**
- Let the point of intersection of two tangents with slopes \( m_1 \) and \( m_2 \) be \( (h, k) \).
- The equations for the tangents can be set up, and by manipulating these equations, we can derive a relationship involving \( h \) and \( k \).

3. **Condition for Perpendicular Tangents:**
- For the tangents to be perpendicular, the product of their slopes must equal \(-1\) (i.e., \( m_1 m_2 = -1 \)).
- The product of the roots of the quadratic equation formed from the tangents will yield the locus of the intersection points.

4. **Conclusion for Statement 2:**
- After simplification, we find that the locus of the point of intersection of the perpendicular tangents is indeed given by:
\[ x^2 + y^2 = a^2 + b^2 \]
- Thus, Statement 2 is true.

### Final Conclusion
Both statements are true. Statement 2 provides a correct explanation for Statement 1.

### Final Answer:
Both statements are true, and Statement 2 is a correct explanation for Statement 1.