A rocket launch pad is located at (15, −7) on a coordinate plane (each unit is 1 meter). For safety, a circular safety zone with a radius of 25 meters is established around the pad. a. Write the general equation representing the boundary of the safety zone. b. What is the circumference of the safety zone? c. What is the area inside the safety zone? d. A worker is standing at point (30,10). Is the worker inside or outside the safety zone?
Answer (1)
Answer:a. General Equation of the Safety Zone Boundary: The general equation of a circle with center (h, k) and radius r is: (x - h)^2 + (y - k)^2 = r^2 In this case, the center (h, k) is (15, -7) and the radius r is 25 meters. Therefore, the equation representing the boundary of the safety zone is: (x - 15)^2 + (y + 7)^2 = 25^2 (x - 15)^2 + (y + 7)^2 = 625 b. Circumference of the Safety Zone: The circumference C of a circle with radius r is given by: C = 2πr With r = 25 meters: C = 2π(25) = 50π meters Approximately: C ≈ 157.08 meters c. Area Inside the Safety Zone: The area A of a circle with radius r is given by: A = πr^2 With r = 25 meters: A = π(25)^2 = 625π square meters Approximately: A ≈ 1963.50 square meters d. Worker's Position Relative to the Safety Zone: To determine if the worker at (30, 10) is inside or outside the safety zone, we substitute the worker's coordinates into the circle's equation: (30 - 15)^2 + (10 + 7)^2 = 15^2 + 17^2 = 225 + 289 = 514 Since 514 < 625, the result of the equation is less than the radius squared. Therefore, the worker is inside the safety zone.
