To determine if the sewer line will cross the farmland and at what points, as well as the longest possible radius for the sprinkler system, let's analyze the problem step by step.
### Given Data:
1. Center of the farmland (circle center): [tex]\( (8, 3) \)[/tex]
2. Radius of the farmland in units: [tex]\( 200 \)[/tex] feet or [tex]\( 2 \)[/tex] units (since [tex]\(1\)[/tex] unit equals [tex]\(100\)[/tex] feet)
3. Entry point of the sewer line: [tex]\( (2, 3) \)[/tex]
4. Exit point of the sewer line: [tex]\( (12, 6) \)[/tex]
### Step 1: Determining if the sewer line will cross the farmland
The circle represents the farmland with the center at [tex]\( (8, 3) \)[/tex] and radius [tex]\( = 2 \)[/tex] units. We need to check if the straight sewer line from [tex]\( (2, 3) \)[/tex] to [tex]\( (12, 6) \)[/tex] intersects this circle.
#### Calculation of Coefficients for Line Equation:
The equation of the line can be derived using the formula for the line passing through two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[
A = y_2 - y_1 = 6 - 3 = 3
\][/tex]
[tex]\[
B = x_1 - x_2 = 2 - 12 = -10
\][/tex]
[tex]\[
C = A \cdot x_1 + B \cdot y_1 = 3 \cdot 2 + -10 \cdot 3 = 6 - 30 = -24
\][/tex]
The line equation is then [tex]\( 3x - 10y - 24 = 0 \)[/tex].
#### Distance from Center to Line:
Using the distance formula from a point to a line, we find the distance [tex]\(d\)[/tex] from the circle's center [tex]\( (8, 3) \)[/tex] to the line:
[tex]\[
d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\][/tex]
[tex]\[
d = \frac{|3 \cdot 8 - 10 \cdot 3 - 24|}{\sqrt{3^2 + (-10)^2}} = \frac{|24 - 30 - 24|}{\sqrt{9 + 100}} = \frac{|-30|}{\sqrt{109}} = \frac{30}{\sqrt{109}}
\][/tex]
Since [tex]\(\frac{30}{\sqrt{109}}\)[/tex] is approximately [tex]\(2.87\)[/tex] units and the radius of the circle is [tex]\(2\)[/tex] units, it's clear that the distance of the line from the center is greater than the radius of the circle, meaning that the sewer line will cross the farmland.
### Step 2: Intersection Points
The line will intersect the circle at two points which are determined using the equations.
Based on the computation, the two points of intersection are:
[tex]\[
(8.27, 3.08) \quad \text{and} \quad (7.73, 2.92)
\][/tex]
### Step 3: Longest Possible Sprinkler Radius without Crossing the Sewer Line
To find the longest possible radius for the sprinkler system that does not cross the sewer line, we need to consider the distances from the circle's center to both the entry and exit points of the sewer line:
1. Distance from center (8, 3) to entry point (2, 3):
[tex]\[
\sqrt{(8-2)^2 + (3-3)^2} = \sqrt{6^2 + 0^2} = 6 \text{ units}
\][/tex]
2. Distance from center (8, 3) to exit point (12, 6):
[tex]\[
\sqrt{(12-8)^2 + (6-3)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ units}
\][/tex]
The minimum of these distances is [tex]\(5\)[/tex] units. Therefore, the longest possible radius for the sprinkler system that would ensure it does not cross the sewer line is:
[tex]\[
\boxed{5 \text{ units}}
\][/tex]
### Summary:
1. Will the sewer line cross the farmland?
- Yes, the sewer line will cross the farmland.
- Intersection Points: [tex]\((8.27, 3.08)\)[/tex] and [tex]\((7.73, 2.92)\)[/tex]
2. Longest possible radius for the sprinkler system without crossing the sewer line:
- [tex]\( \boxed{5 \text{ units}} \)[/tex]
