find the area of the quadrilateral formed by a pair of tangents from the point (4,5) to the circle `x^2 + y^2 -4x -2y-11 = 0` and pair of its radii.

Options

A. `4 sq. units`
B. `8 sq. units`
C. `6 sq. units`
D. `16 sq. units`

Explanation

To find the area of the quadrilateral formed by the pair of tangents from the point (4, 5) to the circle given by the equation \(x^2 + y^2 - 4x - 2y - 11 = 0\) and the pair of its radii, we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form.
The given equation of the circle is:
\[ x^2 + y^2 - 4x - 2y - 11 = 0 \]
We can rearrange it to complete the square:
\[ (x^2 - 4x) + (y^2 - 2y) = 11 \]
Completing the square for \(x\) and \(y\):
\[ (x - 2)^2 - 4 + (y - 1)^2 - 1 = 11 \]
\[ (x - 2)^2 + (y - 1)^2 = 16 \]
Thus, the center of the circle \(C\) is at \((2, 1)\) and the radius \(r\) is \(4\) (since \(\sqrt{16} = 4\)).

### Step 2: Find the distance from the point (4, 5) to the center of the circle.
Let \(P(4, 5)\) be the external point. The distance \(PC\) from point \(P\) to the center \(C(2, 1)\) is calculated as follows:
\[ PC = \sqrt{(4 - 2)^2 + (5 - 1)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

### Step 3: Calculate the length of the tangents from point \(P\) to the circle.
The length of the tangent \(AP\) from point \(P\) to the circle can be found using the formula:
\[ AP = \sqrt{PC^2 - r^2} \]
Substituting the values:
\[ AP = \sqrt{(2\sqrt{5})^2 - 4^2} = \sqrt{20 - 16} = \sqrt{4} = 2 \]
Since there are two tangents from point \(P\), \(AP\) and \(BP\) are both equal to \(2\).

### Step 4: Calculate the area of triangle \(ABC\).
The area of triangle \(ABC\) formed by the tangents and the radius can be calculated using the formula:
\[ \text{Area}_{\triangle ABC} = \frac{1}{2} \times AP \times AC \]
Where \(AC\) is the radius of the circle, which is \(4\):
\[ \text{Area}_{\triangle ABC} = \frac{1}{2} \times 2 \times 4 = 4 \text{ square units} \]

### Step 5: Calculate the area of the quadrilateral \(APBC\).
The quadrilateral \(APBC\) consists of two triangles \(APC\) and \(BPC\). Since both triangles are congruent, the area of quadrilateral \(APBC\) is:
\[ \text{Area}_{APBC} = 2 \times \text{Area}_{\triangle ABC} = 2 \times 4 = 8 \text{ square units} \]

### Final Answer:
The area of the quadrilateral formed by the pair of tangents from the point (4, 5) to the circle and the pair of its radii is:
\[ \boxed{8} \text{ square units} \]