If `S -= x ^(2) + y ^(2) - 2x - 4y - 4=0, L -= 2x + 2y + 15=0 and P (3,4)` represent a circle, a line and a pont respectively then

Options

A. L is tangent to S to P
B. L is polar of P with respect to S
C. L is the chord of contact of P with respect to S
D. P is inside and L is outside S

Answer

To solve the problem step by step, we need to analyze the given equations and find the relationship between the circle, the line, and the point.

### Step 1: Write the equation of the circle in standard form
The given equation of the circle is:
\[ S: x^2 + y^2 - 2x - 4y - 4 = 0 \]

We can rearrange this equation to identify the center and radius. We will complete the square for both \(x\) and \(y\).

1. Rearranging the equation:
\[ x^2 - 2x + y^2 - 4y = 4 \]

2. Completing the square for \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

3. Completing the square for \(y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

4. Substitute these back into the equation:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = 4 \]
\[ (x - 1)^2 + (y - 2)^2 - 5 = 4 \]
\[ (x - 1)^2 + (y - 2)^2 = 9 \]

Now, we can identify the center and radius:
- Center \(C(1, 2)\)
- Radius \(R = \sqrt{9} = 3\)

### Step 2: Write the equation of the line
The given equation of the line is:
\[ L: 2x + 2y + 15 = 0 \]

We can rearrange this to find the slope-intercept form:
\[ 2y = -2x - 15 \]
\[ y = -x - \frac{15}{2} \]

### Step 3: Find the perpendicular distance from the center of the circle to the line
To find the distance \(d\) from the center \(C(1, 2)\) to the line \(L: Ax + By + C = 0\) where \(A = 2\), \(B = 2\), and \(C = 15\), we use the formula:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \(x_0 = 1\) and \(y_0 = 2\):
\[ d = \frac{|2(1) + 2(2) + 15|}{\sqrt{2^2 + 2^2}} = \frac{|2 + 4 + 15|}{\sqrt{4 + 4}} = \frac{|21|}{\sqrt{8}} = \frac{21}{2\sqrt{2}} = \frac{21\sqrt{2}}{4} \]

### Step 4: Compare the distance with the radius
Now we compare the distance \(d\) with the radius \(R\):
- Radius \(R = 3\)
- Distance \(d = \frac{21\sqrt{2}}{4}\)

To determine if the line is inside, outside, or tangent to the circle, we need to compare \(d\) with \(R\):
1. Calculate \(R^2 = 9\)
2. Calculate \(d^2 = \left(\frac{21\sqrt{2}}{4}\right)^2 = \frac{441 \cdot 2}{16} = \frac{882}{16} = 55.125\)

Since \(d^2 > R^2\), the line is outside the circle.

### Step 5: Determine the position of point \(P(3, 4)\)
To check if point \(P(3, 4)\) is inside or outside the circle, we calculate the distance from the center \(C(1, 2)\) to point \(P(3, 4)\):
\[ d_P = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \]

Now, we compare \(d_P\) with \(R\):
- \(d_P^2 = (2\sqrt{2})^2 = 8\)
- \(R^2 = 9\)

Since \(d_P^2 < R^2\), point \(P\) is inside the circle.

### Conclusion
- The line \(L\) is outside the circle \(S\).
- The point \(P(3, 4)\) is inside the circle \(S\).