Drone Path and a Circular No-Fly Zone
On a coordinate map, units are 100 m. A circular no-fly zone is centred at C(4, -1) with radius 5 units. A drone is programmed to fly along the straight line L with equation 3x - 4y + 7 = 0. The operator claims that because the line does not pass through the centre, the drone will not enter the zone. A safety officer wants to know whether the path intersects the zone, and if so, how long the drone would be inside the zone. A second, parallel flight path is being considered with equation 3x - 4y + k = 0. The safety officer wants the smallest positive value of k for which the new path just touches the boundary of the zone, and needs to know the coordinates of the point of contact. All coordinates are in the same units.
1. Determine whether the drone's path intersects the no-fly zone. Justify your answer without drawing a diagram.
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The zone is the circle (x-4)²+(y+1)²=25, with centre C(4,-1) and radius r=5.
The distance from C to line L: 3x-4y+7=0 is d=|3(4)-4(-1)+7|/√(3²+(-4)²)=|12+4+7|/5=23/5=4.6.
Since d=4.6<5=r, the line is closer to the centre than the radius, so it cuts the circle at two distinct points. The operator's reasoning is invalid: a line that does not pass through the centre can still intersect the circle. The drone's path does enter the no-fly zone.
2. Find the length of the path inside the zone, in metres.
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For a chord at perpendicular distance d from the centre, half the chord length is √(r²-d²). Here r=5 and d=4.6, so half-chord = √(25-21.16)=√3.84≈1.9596. Full chord = 2×1.9596≈3.9192 units.
Each unit is 100 m, so the path inside the zone is about 391.9 m, or 392 m to 3 significant figures. This is the length of the intersection of line L with the circular zone.
3. Find the smallest positive value of k for which the parallel path 3x-4y+k=0 is tangent to the zone, and give the point of contact.
Read the worked answer
Tangency means the distance from C(4,-1) to the line equals r=5: |3(4)-4(-1)+k|/5=5 |16+k|=25 k=9 or k=-41. The smallest positive value is k=9.
At tangency the point of contact is the foot of the perpendicular from C to the line. For k=9, the line is 3x-4y+9=0. The foot is C - (3(4)-4(-1)+9)/(3²+(-4)²) = (4,-1) - (25/25)(3,-4) = (4-3, -1+4) = (1,3).
Check: (1-4)²+(3+1)²=9+16=25, so (1,3) is on the circle, and 3(1)-4(3)+9=0, so it is on the line. Thus the tangent path just touches the zone at (1,3).