`S:x ^(2) + y ^(2) -6x + 4y -3=0` is a circle and `L : 4x + 3y + 19=0` is a straight line.

Options

A. L is a chord of S.
B. L is a diameter of S
C. L is a tangent to S
D. none of these

Answer

To determine the relationship between the given circle and the line, we will follow these steps:

### Step 1: Identify the Circle's Equation
The equation of the circle is given as:
\[ S: x^2 + y^2 - 6x + 4y - 3 = 0 \]

### Step 2: Rewrite the Circle's Equation in Standard Form
To find the center and radius of the circle, we need to rewrite the equation in standard form \((x - h)^2 + (y - k)^2 = r^2\).

1. Group the \(x\) and \(y\) terms:
\[ (x^2 - 6x) + (y^2 + 4y) = 3 \]

2. Complete the square for \(x\):
- Take half of the coefficient of \(x\) (which is -6), square it: \((-6/2)^2 = 9\).
- Add and subtract 9:
\[ (x^2 - 6x + 9 - 9) = (x - 3)^2 - 9 \]

3. Complete the square for \(y\):
- Take half of the coefficient of \(y\) (which is 4), square it: \((4/2)^2 = 4\).
- Add and subtract 4:
\[ (y^2 + 4y + 4 - 4) = (y + 2)^2 - 4 \]

4. Substitute back into the equation:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 = 3 \]
\[ (x - 3)^2 + (y + 2)^2 - 13 = 3 \]
\[ (x - 3)^2 + (y + 2)^2 = 16 \]

### Step 3: Identify the Center and Radius
From the standard form \((x - 3)^2 + (y + 2)^2 = 16\):
- Center \((h, k) = (3, -2)\)
- Radius \(r = \sqrt{16} = 4\)

### Step 4: Identify the Line's Equation
The equation of the line is given as:
\[ L: 4x + 3y + 19 = 0 \]

### Step 5: Calculate the Perpendicular Distance from the Center to the Line
The formula for the distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Substituting \(A = 4\), \(B = 3\), \(C = 19\), and the center \((x_1, y_1) = (3, -2)\):
\[ d = \frac{|4(3) + 3(-2) + 19|}{\sqrt{4^2 + 3^2}} \]
\[ = \frac{|12 - 6 + 19|}{\sqrt{16 + 9}} \]
\[ = \frac{|25|}{\sqrt{25}} = \frac{25}{5} = 5 \]

### Step 6: Compare the Distance with the Radius
- Radius of the circle \(r = 4\)
- Distance from the center to the line \(d = 5\)

Since the distance \(d\) (5) is greater than the radius \(r\) (4), the line does not intersect the circle.

### Conclusion
The line does not touch or intersect the circle, meaning it is located outside the circle.

---