Application: Pole on Circular Boundary
Apply quadratic equations to solve real-world geometry problems involving circles and distances.
Ch. 4 Quadratic Equations Applications Hard
Instructions
- Read the problem: A pole must be erected on the boundary of a circular park with two diametrically opposite gates.
- Apply Thales' theorem: When P is on a circle with diameter AB, then ∠APB = 90°, so PA² + PB² = AB².
- Set up equations using the given difference of distances.
- Form and solve the quadratic equation.
- Check if the solution is feasible (real and positive distances).
- Interpret the solution in the context of the problem.
Problem
Generating problem...
Progress 0/7 steps completed
Step 1 — Identify given information
Extract the diameter and distance difference from the problem.
Diameter =
Enter the park diameter value from the problem statement
meters Distance difference =
Enter the distance difference value from the problem statement
meters
Look for the diameter and distance difference values in the problem
statement above.
Step 2 — Apply Thales' Theorem
Since P is on the circle with diameter AB, we have PA² + PB² = AB².
Calculate AB² (diameter²).
PA² + PB² =
Calculate the square of the diameter
Square the diameter value. If diameter = d, then d² = d × d.
Step 3 — Set up the system of equations
Let PA = x and PB = y. Write the two equations.
Equation 1 (difference): x - y = ___
Equation 2 (Pythagoras): x² + y² =
___
The first equation comes from the distance difference: PA - PB =
difference. The second equation comes from Thales' theorem: PA² +
PB² = diameter².
Step 4 — Form quadratic equation in y
Substitute x = y + difference into equation 2 to get: 2y² + ay + b =
0
2y² +
Enter the coefficient of the linear term in the quadratic equation
y +
Enter the constant coefficient in the quadratic equation
= 0
Substitute x = y + difference into x² + y² = d². Expand (y +
difference)² = y² + 2(difference)y + (difference)². Combine like
terms.
Hint: Expand (y + Δ)² + y² = d²
Step 5 — Check feasibility (discriminant)
Calculate discriminant D = a² - 4(2)(b) to check if real solutions
exist.
D =
Calculate the discriminant to determine if real solutions exist
For equation 2y² + ay + b = 0, the discriminant is D = a² - 4(2)(b)
= a² - 8b. If D ≥ 0, real solutions exist.
Step 6 — Solve for distances
Use quadratic formula to find y, then calculate x = y + difference.
Distance 1 (PA) =
Enter the calculated distance from pole to gate A
meters Distance 2 (PB) =
Enter the calculated distance from pole to gate B
meters
Use the quadratic formula: y = (-a ± √D) / 4. Choose the positive
solution. Then x = y + difference.
Step 7 — Final Answer
Based on your solution, is it possible to erect the pole?
Choose whether it is possible to erect the pole based on your
calculations
If the discriminant is positive and you found real, positive
distances, then it's possible. If the discriminant is negative, then
it's not possible.
Concept Mastery
Distance Relations
Thales' Theorem
Quadratic Setup
Discriminant
Visualization (GeoGebra)
Key Concepts
Thales' Theorem
If P is any point on a circle with diameter AB, then angle ∠APB =
90°
This means PA² + PB² = AB² (Pythagorean theorem)
System of Equations
Combine the distance difference with the Pythagorean relation to
form a quadratic equation
Feasibility Check
The discriminant tells us if a solution exists in the real world
Why This Matters
This problem demonstrates how quadratic equations appear in real-world scenarios involving:
- Urban planning and park design
- Engineering constraint optimization
- Geometric problem-solving in architecture
- GPS and navigation systems using distance calculations
Learning Connection
This activity bridges algebra (quadratic equations) with geometry
(circles and distances), showing how mathematical concepts work
together to solve practical problems.
Challenge Level
Select problem difficulty level
Choose your challenge level before generating a new problem.