From the point P(3, 4) tangents PA and PB are drawn to the circle `x^(2)+y^(2)+4x+6y-12=0`. The area of `Delta` PAB in square units, is

Options

A. `(1323)/(42)`
B. `(1715)/(74)`
C. `(926)/(17)`
D. `(1409)/(13)`

Explanation

To find the area of triangle PAB formed by the tangents PA and PB drawn from point P(3, 4) to the circle given by the equation \(x^2 + y^2 + 4x + 6y - 12 = 0\), we will follow these steps:

### Step 1: Find the center and radius of the circle
The equation of the circle is given as:
\[ x^2 + y^2 + 4x + 6y - 12 = 0 \]
We can rewrite this in the standard form \((x - h)^2 + (y - k)^2 = r^2\).

To find the center \((h, k)\), we complete the square for \(x\) and \(y\):
- For \(x\): \(x^2 + 4x\) can be rewritten as \((x + 2)^2 - 4\).
- For \(y\): \(y^2 + 6y\) can be rewritten as \((y + 3)^2 - 9\).

Substituting these back into the equation gives:
\[ (x + 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0 \]
Simplifying this, we have:
\[ (x + 2)^2 + (y + 3)^2 - 25 = 0 \]
Thus, the center of the circle is \((-2, -3)\) and the radius \(r\) is:
\[ r = \sqrt{25} = 5 \]

### Step 2: Calculate the length of the tangents PA and PB
The length of the tangent from point \(P(3, 4)\) to the circle can be calculated using the formula:
\[ PA = PB = \sqrt{S_1} \]
where \(S_1\) is obtained by substituting the coordinates of point \(P\) into the circle's equation:
\[ S_1 = x^2 + y^2 + 4x + 6y - 12 \]
Substituting \(P(3, 4)\):
\[ S_1 = 3^2 + 4^2 + 4(3) + 6(4) - 12 \]
Calculating this gives:
\[ S_1 = 9 + 16 + 12 + 24 - 12 = 49 \]
Thus, the length of the tangent is:
\[ PA = PB = \sqrt{49} = 7 \]

### Step 3: Find the angle \(\theta\) at point \(P\)
Using the triangle PAC, where \(C\) is the center of the circle, we can find \(\tan \theta\):
\[ \tan \theta = \frac{CA}{PA} = \frac{5}{7} \]

### Step 4: Calculate \(\sin 2\theta\)
Using the double angle formula:
\[ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \]
Substituting \(\tan \theta = \frac{5}{7}\):
\[ \sin 2\theta = \frac{2 \cdot \frac{5}{7}}{1 + \left(\frac{5}{7}\right)^2} = \frac{\frac{10}{7}}{1 + \frac{25}{49}} = \frac{\frac{10}{7}}{\frac{74}{49}} = \frac{10 \cdot 49}{7 \cdot 74} = \frac{490}{518} = \frac{70}{74} \]

### Step 5: Calculate the area of triangle PAB
The area \(A\) of triangle PAB can be calculated using:
\[ A = \frac{1}{2} \cdot PA \cdot PB \cdot \sin 2\theta \]
Substituting the values:
\[ A = \frac{1}{2} \cdot 7 \cdot 7 \cdot \frac{70}{74} = \frac{49 \cdot 70}{148} = \frac{3430}{148} \]

### Final Result
Thus, the area of triangle PAB is:
\[ A = \frac{1715}{74} \text{ square units} \]