What is the area of the quadrilateral formed by a pair of tangents from the point (4, 5) to the circle x² + y² - 4x - 2y - 11 = 0 and a pair of its radii?

Steps to Solve

1. Rewrite the Equation of the Circle

First, we need to rewrite the equation of the circle in standard form.

The given equation is: $$ x^2 + y^2 - 4x - 2y - 11 = 0 $$

Rearranging gives: $$ x^2 - 4x + y^2 - 2y = 11 $$

Next, complete the square for both $x$ and $y$.

2. Complete the Square

For $x^2 - 4x$, complete the square: $$ x^2 - 4x = (x - 2)^2 - 4 $$

For $y^2 - 2y$, complete the square: $$ y^2 - 2y = (y - 1)^2 - 1 $$

Substituting these into the equation gives: $$ (x - 2)^2 - 4 + (y - 1)^2 - 1 = 11 $$

Simplifying leads to: $$ (x - 2)^2 + (y - 1)^2 = 16 $$

This indicates a circle centered at $(2, 1)$ with a radius of $4$.

3. Calculate the Distance from Point to Center

Next, calculate the distance from the point $(4, 5)$ to the center of the circle $(2, 1)$: $$ d = \sqrt{(4 - 2)^2 + (5 - 1)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} $$

4. Use the Tangent Length Formula

The length of each tangent from point $(4, 5)$ to the circle is given by: $$ \text{Tangent Length} = \sqrt{d^2 - r^2} $$

Where $d$ is the distance calculated, and $r$ is the radius of the circle: $$ \text{Tangent Length} = \sqrt{(2\sqrt{5})^2 - 4^2} = \sqrt{20 - 16} = \sqrt{4} = 2 $$

5. Find the Area of the Quadrilateral

The area of the quadrilateral formed by the tangents and the radius is: $$ \text{Area} = \frac{1}{2} \times \text{Tangent Length}^2 \times \sin 90^\circ = \frac{1}{2} \times 2^2 = \frac{1}{2} \times 4 = 2 $$

But since the quadrilateral has two tangents forming two identical triangles, we multiply by $2$: $$ \text{Total Area} = 2 \times 2 = 4 \text{ sq-units} $$