8 TePR2 p? 4x- 2) 11 = 0 be a circle. A pall or tangents from the point (4, 5) with a pair of radii form a quadrilateral of area (a) 8 sq. units (b) 10 sq. units (d) 16 sq. units (c) 12 sq. units

Answer

To solve the problem, we need to find the area of the quadrilateral formed by the tangents from the point (4, 5) to the circle defined by the equation \(x^2 + y^2 - 4x - 2y + 11 = 0\). Let's break down the solution step by step.

### Step 1: Rewrite the Circle Equation
First, we rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 4x - 2y + 11 = 0 \]
We can rearrange it as:
\[ x^2 - 4x + y^2 - 2y = -11 \]
Next, we complete the square for the \(x\) and \(y\) terms.

### Step 2: Completing the Square
For \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
For \(y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Now substituting back into the equation:
\[ (x - 2)^2 - 4 + (y - 1)^2 - 1 = -11 \]
This simplifies to:
\[ (x - 2)^2 + (y - 1)^2 = -11 + 5 = -6 \]
This indicates an error in the original equation since the radius cannot be imaginary. Let's correct the constant term:
\[ (x - 2)^2 + (y - 1)^2 = 16 \quad \text{(after correcting the constant)} \]
Thus, the center of the circle is \(C(2, 1)\) and the radius \(r = 4\).

### Step 3: Calculate the Distance from Point to Center
Now, we calculate the distance \(PC\) from point \(P(4, 5)\) to the center \(C(2, 1)\):
\[ PC = \sqrt{(4 - 2)^2 + (5 - 1)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

### Step 4: Use Pythagorean Theorem
Since \(PC\) is the hypotenuse of the right triangle formed with the radius and the tangent, we can find the length of the tangent \(PT\) using the Pythagorean theorem:
\[ PC^2 = PT^2 + r^2 \]
Substituting the known values:
\[ (2\sqrt{5})^2 = PT^2 + 4^2 \]
\[ 20 = PT^2 + 16 \]
\[ PT^2 = 20 - 16 = 4 \implies PT = 2 \]

### Step 5: Area of the Quadrilateral
The area of the quadrilateral formed by the two tangents and the radii can be calculated as follows:
1. The area of triangle \(PQC\) is given by:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Where the base \(PQ = 2\) and height (the radius) = \(4\):
\[ \text{Area of } \triangle PQC = \frac{1}{2} \times 2 \times 4 = 4 \]
2. Since there are two such triangles, the total area of the quadrilateral \(QCRP\) is:
\[ \text{Total Area} = 2 \times 4 = 8 \text{ sq. units} \]

### Final Answer
Thus, the area of the quadrilateral formed by the tangents and the radii is:
\[ \boxed{8 \text{ sq. units}} \]