`x^(2)+y^(2)-4x-2y-11=0` is a circle to which tangents are drawn from the point (4, 5) which form a quadrilateral with a pair of radii. The area of this quadrilateral in sq. units is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we start by rewriting the equation of the circle and identifying its center and radius. The equation given is:

\[ x^2 + y^2 - 4x - 2y - 11 = 0 \]

### Step 1: Rewrite the equation in standard form

We can complete the square for \(x\) and \(y\):

1. Rearranging the equation:
\[ (x^2 - 4x) + (y^2 - 2y) = 11 \]

2. Completing the square:
- For \(x\): \(x^2 - 4x = (x - 2)^2 - 4\)
- For \(y\): \(y^2 - 2y = (y - 1)^2 - 1\)

Substituting back, we get:
\[ (x - 2)^2 - 4 + (y - 1)^2 - 1 = 11 \]
\[ (x - 2)^2 + (y - 1)^2 = 16 \]

### Step 2: Identify the center and radius of the circle

From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we find:
- Center \(C(2, 1)\)
- Radius \(r = \sqrt{16} = 4\)

### Step 3: Calculate the length of the tangent from point (4, 5)

Using the formula for the length of the tangent \(L\) from a point \((x_1, y_1)\) to a circle centered at \((h, k)\) with radius \(r\):
\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]

Substituting the values:
- Point \(P(4, 5)\)
- Center \(C(2, 1)\)
- Radius \(r = 4\)

Calculating:
\[ L = \sqrt{(4 - 2)^2 + (5 - 1)^2 - 4^2} \]
\[ = \sqrt{(2)^2 + (4)^2 - 16} \]
\[ = \sqrt{4 + 16 - 16} = \sqrt{4} = 2 \]

### Step 4: Calculate the area of the quadrilateral formed

The quadrilateral \(PABC\) consists of two triangles \(PAC\) and \(PBC\). Since \(PA\) and \(PB\) are tangents from point \(P\) to the circle, they are equal in length, and the angles at point \(C\) are \(90^\circ\).

1. Area of triangle \(PAC\):
\[ \text{Area}_{PAC} = \frac{1}{2} \times PA \times AC \]
Here, \(PA = 2\) (length of tangent) and \(AC = 4\) (radius).
\[ \text{Area}_{PAC} = \frac{1}{2} \times 2 \times 4 = 4 \text{ square units} \]

2. Since \(PBC\) is congruent to \(PAC\):
\[ \text{Area}_{PBC} = 4 \text{ square units} \]

### Step 5: Total area of quadrilateral \(PABC\)

\[ \text{Area}_{PABC} = \text{Area}_{PAC} + \text{Area}_{PBC} = 4 + 4 = 8 \text{ square units} \]

Thus, the area of the quadrilateral formed is:

\[ \boxed{8} \text{ square units} \]