Walkway Tangent to a Circular Park
A city plans a straight elevated walkway across a circular park. On a coordinate map, the park boundary is the circle x² + y² = 25, where units are 10 m. The proposed walkway follows the line y = mx + 6. The project engineer must choose the slope m so that the walkway crosses the park, but a protected wetland occupies the inner circle x² + y² ≤ 16, and the walkway must not enter that inner circle. The engineer initially considers only whether the line meets the outer circle, then realizes the wetland imposes a second condition. The walkway is a straight line with no thickness; safety rules require it to stay at least 4 units from the park centre and at most 5 units from the centre at its closest point. The engineer wants the full range of slopes that satisfies both conditions, plus the tangent case.
1. Find the range of m for which the walkway crosses the outer circle but does not enter the inner circle. Show the distance calculations.
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The line is mx - y + 6 = 0. Its distance from the centre (0,0) is d=6/√(m²+1). Crossing the outer circle requires d≤5: 6/√(m²+1)≤5, so √(m²+1)≥6/5, m²+1≥36/25, m²≥11/25, hence |m|≥√11/5≈0.663. Avoiding the inner wetland requires d≥4: 6/√(m²+1)≥4, so √(m²+1)≤3/2, m²+1≤9/4, m²≤5/4, hence |m|≤√5/2≈1.118. Both conditions together give √11/5≤|m|≤√5/2. The line must be steep enough to reach the outer circle but not so steep that it cuts into the inner circle.
2. For the tangent case m=√11/5, find the point where the line touches the outer circle.
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At m=√11/5 the distance is exactly 5, so the line is tangent to the outer circle. Write the line as (√11/5)x - y +6=0, with a=√11/5, b=-1, c=6. The foot of the perpendicular from the origin is (x0,y0)=(-ac/(a²+b²), -bc/(a²+b²)). Now a²+b²=11/25+1=36/25. Thus x0=-(√11/5·6)/(36/25)=-(6√11/5)(25/36)=-5√11/6, and y0=-(-1·6)/(36/25)=6·25/36=25/6. The tangent point is (-5√11/6, 25/6). Check: x0²+y0²=(25·11)/36+625/36=900/36=25, and substitution into the line gives 25/6.
3. If the protected radius becomes 5 units, describe the allowed slopes. What does this show about the tangent condition?
Read the worked answer
If the protected radius becomes 5 units, the walkway must satisfy d≥5 as well as d≤5, so d=5 exactly. Then 6/√(m²+1)=5, giving m²=11/25 and m=±√11/5. These are the two tangent lines to the outer circle; they touch the boundary but do not enter the protected region. If the protected radius were slightly less than 5, a narrow interval of slopes around each tangent value would remain, but at radius 5 the interval collapses to two isolated slopes. This shows how a tangent condition arises as a limiting case of an inequality.