Let A,B, and C be three sets such that
`A={(x,y)|(x)/(cos theta)=(y)/(sintheta)=5,"where" 'theta'"is parameter"}`
`B= {(x,y)|(x-3)/(cos phi)=(y-4)/(sin phi)=r}`
`C= { (x,y)|(x-3)^(2)+(y-4)^(2)leR^(2)}`
If `phi` is fixed and r varies and `(A cap B) =1`, then `sec phi` is equal to (a) `(5)/(4)` (b) `(-5)/(4)` (c) `(5)/(3)` (d) `(-5)/(3)`

Options

A. `(5)/(4)`
B. `(-5)/(4)`
C. `(5)/(3)`
D. `(-5)/(3)`

Explanation

To solve the problem, we need to analyze the sets A, B, and C and their intersections.

### Step 1: Analyze Set A
Set A is defined as:
\[ A = \{(x,y) | \frac{x}{\cos \theta} = \frac{y}{\sin \theta} = 5\} \]

From this, we can express \( x \) and \( y \):
\[ x = 5 \cos \theta \quad \text{and} \quad y = 5 \sin \theta \]

This indicates that set A represents a circle with a radius of 5 centered at the origin (0, 0):
\[ x^2 + y^2 = 25 \]

### Step 2: Analyze Set B
Set B is defined as:
\[ B = \{(x,y) | \frac{x-3}{\cos \phi} = \frac{y-4}{\sin \phi} = r\} \]

From this, we can express \( x \) and \( y \):
\[ x - 3 = r \cos \phi \quad \text{and} \quad y - 4 = r \sin \phi \]
Thus,
\[ x = 3 + r \cos \phi \quad \text{and} \quad y = 4 + r \sin \phi \]

This indicates that set B represents a line parameterized by \( r \) that passes through the point (3, 4) with direction determined by \( \cos \phi \) and \( \sin \phi \).

### Step 3: Analyze Set C
Set C is defined as:
\[ C = \{(x,y) | (x-3)^2 + (y-4)^2 \leq r^2\} \]

This indicates that set C represents a circle centered at (3, 4) with radius \( r \).

### Step 4: Intersection of Sets A and B
Given that \( A \cap B = 1 \), this means that the circle defined by set A touches the line defined by set B at exactly one point.

### Step 5: Condition for Tangency
For the circle \( x^2 + y^2 = 25 \) to touch the line defined by set B, the distance from the center of the circle (0, 0) to the line must equal the radius of the circle (5).

The line can be rewritten in the standard form:
\[ y - 4 = \tan \phi (x - 3) \]
or in the form \( Ax + By + C = 0 \):
\[ \tan \phi \cdot x - y + (4 - 3 \tan \phi) = 0 \]

Where \( A = \tan \phi \), \( B = -1 \), and \( C = 4 - 3 \tan \phi \).

### Step 6: Calculate the Distance from the Origin to the Line
The distance \( d \) from the point (0, 0) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|4 - 3 \tan \phi|}{\sqrt{\tan^2 \phi + 1}} = \frac{|4 - 3 \tan \phi|}{\sec \phi} \]

### Step 7: Set the Distance Equal to the Radius
Setting this distance equal to the radius of the circle (5):
\[ \frac{|4 - 3 \tan \phi|}{\sec \phi} = 5 \]

Multiplying both sides by \( \sec \phi \):
\[ |4 - 3 \tan \phi| = 5 \sec \phi \]

### Step 8: Solve for \( \sec \phi \)
This leads to two cases:
1. \( 4 - 3 \tan \phi = 5 \sec \phi \)
2. \( 4 - 3 \tan \phi = -5 \sec \phi \)

From the first case:
\[ 4 - 3 \tan \phi = 5 \sec \phi \implies 4 = 5 \sec \phi + 3 \tan \phi \]

From the second case:
\[ 4 - 3 \tan \phi = -5 \sec \phi \implies 4 + 5 \sec \phi = 3 \tan \phi \]

Solving these equations will yield the value of \( \sec \phi \).

### Final Step: Evaluate the Options
After solving the equations, we find that the value of \( \sec \phi \) is:
\[ \sec \phi = \frac{5}{4} \]

Thus, the answer is:
**(a) \( \frac{5}{4} \)**