`S: x ^(2) + y ^(2) + 6x - 14y-6 =0` is a circle and
`L: 7x + 3y + 58 =0` is a straight line

Options

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

Answer

To determine the relationship between the circle given by the equation \( S: x^2 + y^2 + 6x - 14y - 6 = 0 \) and the line given by \( L: 7x + 3y + 58 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
We start with the circle's equation:
\[ x^2 + y^2 + 6x - 14y - 6 = 0 \]
We can rewrite this in the standard form by completing the square.

### Step 2: Complete the Square for the Circle
1. For \( x^2 + 6x \):
- Take half of 6, which is 3, and square it to get 9.
- Thus, \( x^2 + 6x = (x + 3)^2 - 9 \).

2. For \( y^2 - 14y \):
- Take half of -14, which is -7, and square it to get 49.
- Thus, \( y^2 - 14y = (y - 7)^2 - 49 \).

Now substituting back into the equation:
\[ (x + 3)^2 - 9 + (y - 7)^2 - 49 - 6 = 0 \]
This simplifies to:
\[ (x + 3)^2 + (y - 7)^2 - 64 = 0 \]
Thus, we have:
\[ (x + 3)^2 + (y - 7)^2 = 64 \]
This shows that the center of the circle is at \( (-3, 7) \) and the radius \( r \) is \( \sqrt{64} = 8 \).

### Step 3: Find the Perpendicular Distance from the Center to the Line
The line equation is:
\[ 7x + 3y + 58 = 0 \]
To find the perpendicular distance \( d \) from the center of the circle \( (-3, 7) \) to the line, we use the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
where \( A = 7 \), \( B = 3 \), \( C = 58 \), and \( (x_1, y_1) = (-3, 7) \).

Substituting these values:
\[ d = \frac{|7(-3) + 3(7) + 58|}{\sqrt{7^2 + 3^2}} = \frac{|-21 + 21 + 58|}{\sqrt{49 + 9}} = \frac{|58|}{\sqrt{58}} = \frac{58}{\sqrt{58}} = \sqrt{58} \]

### Step 4: Compare the Radius and the Perpendicular Distance
The radius of the circle is \( 8 \) and the perpendicular distance from the center to the line is \( \sqrt{58} \).

Now we check:
\[ \sqrt{58} \approx 7.61 < 8 \]

### Conclusion
Since the radius \( 8 \) is greater than the perpendicular distance \( \sqrt{58} \), the line intersects the circle at two points. Therefore, the line is a chord of the circle.