Word Problem: Surveillance Camera Installation

Problem Statement:

The crime investigation division of a provincial police office is setting up a surveillance camera at the entrance of a circular park. The park is circular with a diameter of 200 meters. The police department wants to install the camera on a pole located at point $P(60, 80)$, where the center of the park is at the origin $O(0, 0)$. The camera can capture an area within a radius of 100 meters. To maximize coverage, the camera needs to be directed such that it captures as much of the park's entrance area as possible.

Tasks:

1. Determine the equation of the circle representing the park.
2. Determine the equation of the circle representing the area captured by the camera.
3. Check whether the camera's coverage circle intersects with the park boundary and, if so, find the points of intersection.
4. Graph both the park and the camera's coverage area.

Solution:

1. Equation of the Park (Circle):

   The park is a circle centered at the origin $O(0, 0)$ with a radius $R = \frac{200}{2} = 100$ meters.

   The equation of the park (circle) is given by: $x^2 + y^2 = 100^2$ $x^2 + y^2 = 10,000$

2. Equation of the Camera's Coverage Area (Circle):

   The camera's coverage area is a circle centered at $P(60, 80)$ with a radius $r = 100$ meters.

   The equation of the camera's coverage circle is: $(x - 60)^2 + (y - 80)^2 = 100^2$ $(x - 60)^2 + (y - 80)^2 = 10,000$

3. Finding Points of Intersection:

   To find if the camera's coverage circle intersects with the park, we set the two equations equal:

   $x^2 + y^2 = 10,000$ $(x - 60)^2 + (y - 80)^2 = 10,000$

   Expanding and simplifying the second equation: $(x - 60)^2 + (y - 80)^2 = x^2 - 120x + 3600 + y^2 - 160y + 6400$ $(x - 60)^2 + (y - 80)^2 = x^2 + y^2 - 120x - 160y + 10,000$

   Set the two equations equal to each other: $10,000 = x^2 + y^2 - 120x - 160y + 10,000$

   Simplifying further: [ 0 = -120x -