The focal chords of the parabola `y^(2)=16x` which are tangent to the circle of radius r and centre (6, 0) are perpendicular, then the radius r of the circle is

Options

A. units
B. `sqrt2` units
C. 1 units
D. `(1)/(2)` units

Answer

To solve the problem, we need to find the radius \( r \) of the circle centered at \( (6, 0) \) such that the focal chords of the parabola \( y^2 = 16x \) that are tangent to the circle are perpendicular to each other.

### Step-by-Step Solution:

1. **Identify the Parabola's Properties**:
The given parabola is \( y^2 = 16x \). This can be rewritten in the standard form \( y^2 = 4ax \) where \( 4a = 16 \). Thus, \( a = 4 \). The focus of the parabola is at the point \( (4, 0) \).

**Hint**: Understand the properties of the parabola and identify its focus.

2. **Circle's Properties**:
The circle is centered at \( (6, 0) \) with radius \( r \). The equation of the circle can be expressed as:
\[ (x - 6)^2 + y^2 = r^2 \]

**Hint**: Write down the equation of the circle based on its center and radius.

3. **Focal Chords and Tangents**:
For the parabola, the focal chords can be represented by points \( P(t_1) \) and \( Q(t_2) \) where:
\[ P(t_1) = (4 + 4t_1^2, 8t_1) \quad \text{and} \quad Q(t_2) = (4 + 4t_2^2, 8t_2) \]
The slopes of the tangents at these points are given by:
\[ m_1 = \frac{8t_1}{4 + 4t_1^2} \quad \text{and} \quad m_2 = \frac{8t_2}{4 + 4t_2^2} \]

**Hint**: Recall how to find the points on the parabola and their corresponding slopes.

4. **Condition for Perpendicularity**:
The tangents are perpendicular if the product of their slopes is \( -1 \):
\[ m_1 \cdot m_2 = -1 \]

**Hint**: Set up the equation for the product of the slopes to find a relationship between \( t_1 \) and \( t_2 \).

5. **Using Geometry**:
The distance from the center of the circle \( (6, 0) \) to the focus \( (4, 0) \) is:
\[ d = 6 - 4 = 2 \]
The triangle formed by the points \( P, Q, \) and the center of the circle has a right angle at the center. Thus, we can apply the Pythagorean theorem.

6. **Applying Pythagorean Theorem**:
In the right triangle formed:
\[ PR^2 = PQ^2 + QR^2 \]
Where \( PR = 2 \), \( PQ = r \), and \( QR = r \):
\[ 2^2 = r^2 + r^2 \implies 4 = 2r^2 \implies r^2 = 2 \implies r = \sqrt{2} \]

**Hint**: Use the Pythagorean theorem to relate the sides of the triangle.

### Final Answer:
The radius \( r \) of the circle is \( \sqrt{2} \).