Circular Fountain and a Straight Walkway
A circular fountain in a park has centre (3, 4) and radius 2 m; its boundary is x² + y² - 6x - 8y + 21 = 0. A straight walkway is to start at the entrance P(0, 0). The parks committee proposes the line through P and a bench at B(8, 6). A gardener warns that the path will cut the fountain. The committee wants to know all slopes m of lines through P that keep at least 1 m clearance from the fountain edge, meaning the distance from the centre to the line must be at least 3 m. It also wants to know whether the proposed PB satisfies this rule. A second proposal is a tangent path that just touches the fountain; the committee needs the exact slopes for a design drawing. The vertical line x = 0 is also under consideration as a boundary path.
1. Find all slopes m of non-vertical lines through P that keep at least 1 m clearance from the fountain edge. Test the proposed line PB, and state how the vertical line x = 0 fits the rule.
Read the worked answer
A non-vertical line through P has equation y = mx, or mx - y = 0. The distance from the centre (3, 4) to this line is
|3m - 4| / √(m² + 1).
For at least 1 m clearance from the fountain edge, this distance must be at least 2 + 1 = 3:
|3m - 4| / √(m² + 1) ≥ 3.
Squaring both sides gives (3m - 4)² ≥ 9(m² + 1), so 9m² - 24m + 16 ≥ 9m² + 9, hence -24m + 7 ≥ 0, or m ≤ 7/24.
The proposed PB has slope (6 - 0)/(8 - 0) = 3/4 = 0.75, which is greater than 7/24 ≈ 0.292. Its distance from the centre is |3(0.75) - 4| / √(1.75) = 1.75 / 1.25 = 1.4 m, which is less than the radius 2; the path actually cuts the fountain. It fails the clearance rule badly.
The vertical line x = 0 has distance 3 from (3, 4), so it meets the 1 m clearance rule exactly. It is not represented by a finite slope m and must be checked separately.
2. Find the exact slopes of the tangent lines through P. Explain why the clearance rule m ≤ 7/24 is not the same as tangency.
Read the worked answer
Tangency means the distance from the centre to the line equals the radius 2:
|3m - 4| / √(m² + 1) = 2.
Squaring gives (3m - 4)² = 4(m² + 1), so 9m² - 24m + 16 = 4m² + 4, hence 5m² - 24m + 12 = 0.
Using the quadratic formula,
m = [24 ± √(576 - 240)] / 10 = [24 ± √336] / 10 = (12 ± 2√21) / 5.
Numerically these are about 0.567 and 4.233. Both tangent lines pass through P and touch the fountain; they are the limiting directions between cutting and clearing the fountain.
The clearance rule m ≤ 7/24 gives distance at least 3, which is stricter than tangency. Tangency has distance exactly 2, so tangent slopes are not included in the 1 m clearance set unless the committee later reduces the clearance requirement to zero.