Setting Out a Circular Pond and Its Tangent Path
A landscape architect is setting out a circular pond on a surveyed grid measured in metres. Three fixed items must sit exactly on the rim: a fountain base at A(0, 5), a bench at B(3, 6) and a lamp post at C(7, 4). A straight gravel path is to touch the pond at C and continue to a service point D(4, 8), and paving around the rim must meet the chord CA at a stated angle so that the edging strips can be pre-cut. Because the contractor's machine follows only straight lines and circular arcs, the centre, radius and tangent must be supplied as exact values with a short verification, not as a sketch. On site, a checking surveyor measures the interior angle ABC as 135° and challenges the drawing, which annotates the paving angle at C as 45°. The architect must show that the pond through A, B and C is the intended circle, that D really lies on the tangent, and that the two angle figures are not in conflict.
1. Find the centre and radius of the circle through A, B and C, and verify that all three points lie on it.
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The centre lies on the perpendicular bisector of every chord.
Chord AB: midpoint (1.5, 5.5), gradient (6 − 5)/(3 − 0) = 1/3, so the perpendicular gradient is −3. Equation: y − 5.5 = −3(x − 1.5), i.e. y = −3x + 10.
Chord BC: midpoint (5, 5), gradient (4 − 6)/(7 − 3) = −1/2, so the perpendicular gradient is 2. Equation: y − 5 = 2(x − 5), i.e. y = 2x − 5.
Solve simultaneously: −3x + 10 = 2x − 5 gives 5x = 15, x = 3, and then y = 1. Centre (3, 1).
Radius = distance to A = √(3² + 4²) = 5.
Check B: √((3 − 3)² + (6 − 1)²) = √25 = 5. Check C: √((7 − 3)² + (4 − 1)²) = √(16 + 9) = 5. All three concyclic.
Circle: (x − 3)² + (y − 1)² = 25.
2. Find the equation of the tangent at C, show that D(4, 8) lies on it, and verify tangency independently using the distance from the centre.
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The radius to C has gradient (4 − 1)/(7 − 3) = 3/4, so the tangent gradient is −4/3.
Tangent: y − 4 = −(4/3)(x − 7). Multiplying by 3: 3y − 12 = −4x + 28, so 4x + 3y = 40.
Test D(4, 8): 4(4) + 3(8) = 16 + 24 = 40. D lies on the line, so the service point is on the proposed path, at distance CD = √((7 − 4)² + (4 − 8)²) = √(9 + 16) = 5 m from the lamp post.
Independent check of tangency: the distance from the centre (3, 1) to the line 4x + 3y − 40 = 0 is |4(3) + 3(1) − 40|/√(4² + 3²) = |12 + 3 − 40|/5 = 25/5 = 5, which equals the radius. A line at distance equal to the radius from the centre is tangent, so the path touches the pond only at C and does not cut the rim.
3. Using vectors, find the angle between the tangent ray CD and the chord CA, and reconcile the surveyor's 135° with the 45° marked on the drawing.
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Vectors from C: CD = D − C = (−3, 4) and CA = A − C = (−7, 1).
CD · CA = (−3)(−7) + (4)(1) = 21 + 4 = 25; |CD| = 5, |CA| = √50 = 5√2.
cos∠DCA = 25/(5 × 5√2) = 1/√2, so ∠DCA = 45°, matching the paving angle on the drawing.
Surveyor's angle: BA = A − B = (−3, −1) and BC = C − B = (4, −2). BA · BC = −12 + 2 = −10; |BA| = √10, |BC| = √20. cos∠ABC = −10/(√10 × √20) = −10/(10√2) = −1/√2, so ∠ABC = 135°. The measurement is correct.
Reconciliation: the tangent–chord theorem states that the angle between the tangent at C and chord CA equals the angle that chord CA subtends in the alternate segment. The angle subtended at B is 135°, but B and D lie on the same side of chord CA (the cross products (−7)(2) − (1)(−4) = −10 and (−7)(4) − (1)(−3) = −25 have the same sign), so the inscribed angle standing on the other arc is 180° − 135° = 45°. That is exactly ∠DCA. Both figures describe the same geometry: 135° is the interior angle at B, 45° is the tangent–chord angle used for cutting the edging. No conflict exists; the drawing should label which angle it quotes.