If largest and smallest value of `(y-4)/(x-3)` is `p` and `q` where `(x,y)` satisfy `x^(2)+y^(2)-2-6y+9=0` then which of the following is true

Options

A. `p+q=4/3`
B. `q=1`
C. `p=4/3`
D. `pq=4/3`

Explanation

To solve the problem, we need to find the largest and smallest values of the expression \((y-4)/(x-3)\) given that \((x,y)\) satisfies the equation of a circle derived from the equation \(x^2 + y^2 - 2x - 6y + 9 = 0\).

### Step-by-Step Solution:

1. **Rewrite the Circle Equation**:
Start with the given equation:
\[ x^2 + y^2 - 2x - 6y + 9 = 0 \]
Rearranging gives:
\[ x^2 - 2x + y^2 - 6y + 9 = 0 \]
We can complete the square for both \(x\) and \(y\):
\[ (x-1)^2 + (y-3)^2 = 1 \]
This represents a circle centered at \((1, 3)\) with a radius of \(1\).

2. **Parameterize the Circle**:
We can parameterize the circle using trigonometric functions:
\[ x = 1 + \cos(t), \quad y = 3 + \sin(t) \]
where \(t\) ranges from \(0\) to \(2\pi\).

3. **Substitute into the Expression**:
Substitute \(x\) and \(y\) into the expression \((y-4)/(x-3)\):
\[ \frac{y-4}{x-3} = \frac{(3 + \sin(t) - 4)}{(1 + \cos(t) - 3)} = \frac{\sin(t) - 1}{\cos(t) - 2} \]

4. **Find the Maximum and Minimum Values**:
To find the maximum and minimum values of \(\frac{\sin(t) - 1}{\cos(t) - 2}\), we need to analyze the behavior of this function:
- The numerator \(\sin(t) - 1\) varies from \(-1\) (when \(\sin(t) = 0\)) to \(0\) (when \(\sin(t) = 1\)).
- The denominator \(\cos(t) - 2\) varies from \(-1\) (when \(\cos(t) = 1\)) to \(-2\) (when \(\cos(t) = -1\)).
- Thus, the expression will have its largest value when \(\sin(t) = 1\) and \(\cos(t) = -1\), giving us:
\[ \frac{0}{-1} = 0 \]
- The smallest value occurs when \(\sin(t) = 0\) and \(\cos(t) = 1\), giving us:
\[ \frac{-1}{-1} = 1 \]

5. **Identify \(p\) and \(q\)**:
From our analysis, we can identify:
- \(p = 0\) (smallest value)
- \(q = 1\) (largest value)

6. **Conclusion**:
The values of \(p\) and \(q\) are \(0\) and \(1\) respectively. Thus, the relationship we can conclude is:
\[ p + q = 0 + 1 = 1 \]
and \(p \cdot q = 0 \cdot 1 = 0\).