L45 — Circles and Coordinate Geometry — Solutions

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1. Q1 — Equation of a circle

   Fluency
   • (a) x² + y² = 49.
   • (b) (x − 4)² + (y + 1)² = 9.
   • (c) Compare with (x − h)² + (y − k)² = r²: centre (5, −2), radius 4.
   • (d) Centre (0, 0), radius 7.

2. Q2 — Points and circles

   Fluency
   • (a) 6² + 8² = 36 + 64 = 100 = r² ⇒ on the circle.
   • (b) (3−1)² + (4−1)² = 4 + 9 = 13 < 25 ⇒ inside the circle.
   • (c) Where x=0: y=±5. Where y=0: x=±5. Points: (5,0), (−5,0), (0,5), (0,−5).
   • (d) r² = 5² + 12² = 25 + 144 = 169. Equation: x² + y² = 169.

3. Q3 — Tangents to circles

   Fluency
   • (a) Radius from (0,0) to (0,5) is vertical (undefined gradient). Tangent is perpendicular ⇒ horizontal: y = 5.
   • (b) Radius gradient = −3/4. Tangent gradient = 4/3. y + 3 = (4/3)(x − 4) ⇒ 4x − 3y = 25.
   • (c) Centre (2,1). Radius gradient = (2−1)/(5−2) = 1/3. Tangent gradient = −3. y − 2 = −3(x − 5) ⇒ y = −3x + 17.
   • (d) Radius gradient to (5,5) from (0,0) = 1. Tangent gradient = −1. y − 5 = −(x − 5) ⇒ x + y = 10.

4. Q4 — Completing the square

   Fluency
   • (a) (x−2)² − 4 + (y+3)² − 9 + 4 = 0 ⇒ (x−2)² + (y+3)² = 9. Centre (2, −3), radius 3.
   • (b) (x+4)² − 16 + (y−1)² − 1 − 8 = 0 ⇒ (x+4)² + (y−1)² = 25. Centre (−4, 1), radius 5.
   • (c) (x−5)² − 25 + y² = 0 ⇒ (x−5)² + y² = 25. Centre (5, 0), radius 5.
   • (d) (x−1)² − 1 + (y−2)² − 4 + 5 = 0 ⇒ (x−1)² + (y−2)² = 0. Only satisfied when x=1, y=2: degenerate circle — the single point (1, 2).

5. Q5 — Chords and the centre

   Understanding
   • (a) A(3,4): 9+16=25 ✓. B(−3,4): 9+16=25 ✓. Both lie on the circle.
   • (b) M = ((3+(−3))/2, (4+4)/2) = (0, 4). Gradient of AB = (4−4)/(−3−3) = 0 (horizontal chord).
   • (c) Perpendicular to horizontal AB through M(0,4): vertical line x = 0. Substituting x=0 into x=0 passes through (0,0) ✓
   • (d) The centre is equidistant from every point on the circle, so it is equidistant from both endpoints of any chord. The perpendicular bisector is the locus of all such equidistant points, so the centre always lies on it.

6. Q6 — Intersections of circles and lines

   Understanding
   • (a) Sub y=x+1: x²+(x+1)²=25 ⇒ 2x²+2x−24=0 ⇒ (x+4)(x−3)=0. Points: (−4, −3) and (3, 4).
   • (b) Distance from centre (0,0) to y=4 is 4. Radius = 3. Since 4 > 3, the line is outside the circle: 0 intersections.
   • (c) Sub y=3x−10: x²+(3x−10)²=10 ⇒ 10x²−60x+90=0 ⇒ (x−3)²=0. One solution (x=3): tangent.
   • (d) Sub y=x: (x−2)²+x²=8 ⇒ 2x²−4x−4=0 ⇒ x=1±√3. Chord length = √[(2√3)²+(2√3)²] = √24 = 2√6.

7. Q7 — Circle through given points

   Understanding
   • (a) Perp bisector of AB(0,0)–(6,0): x=3. Perp bisector of AC(0,0)–(0,8): y=4. Centre (3, 4), radius = √(9+16) = 5.
   • (b) r = √[(5−2)²+(7−3)²] = √(9+16) = 5. Equation: (x−2)² + (y−3)² = 25.
   • (c) Centre = midpoint of PQ = (4, 3). Radius = 3. Equation: (x−4)² + (y−3)² = 9.
   • (d) r = √(9+16) = 5. Equation: (x−3)² + (y−4)² = 25.

8. Q8 — Mixed coordinate geometry

   Understanding
   • (a) Sub y=2: (x−1)²=9 ⇒ x=4 or x=−2. Chord from (−2,2) to (4,2): length = 6.
   • (b) Using ax+by=r² with a=2, b=3, r²=13: 2x + 3y = 13.
   • (c) d = |2(3)−1−4|/√5 = 1/√5 = √5/5 ≈ 0.45. Radius = √20 = 2√5 ≈ 4.47. Since d < r, the line intersects the circle (not a tangent).
   • (d) Centres: (0,0) and (5,0). Distance = 5. r⊂1+r⊂2; = 2+3 = 5. Since d = r⊂1;+r⊂2;, the circles are externally tangent — they just touch, no overlap.

9. Q9 — Circle in context

   Problem Solving
   • (a) r = |8−3| = 5 units. Circumference = 2π×5×100 = 1000π ≈ 3142 m.
   • (b) (x−4)² + (y−3)² = 25.
   • (c) Sub y=6: (x−4)²+9=25 ⇒ (x−4)²=16 ⇒ x=8 or x=0. Points: (0, 6) and (8, 6).
   • (d) Radius CA from C(4,3) to A(4,8) is vertical. Tangent at A is perpendicular ⇒ horizontal: y = 8.

10. Q10 — Coordinate geometry proof

    Problem Solving
    • (a) AB gradient = 0 (horizontal). BC gradient = undefined (vertical). Horizontal ⊥ vertical ⇒ right angle at B ✓
    • (b) Hypotenuse AC from A(0,0) to C(10,10). Centre = (5, 5). Radius = √(25+25) = 5√2.
    • (c) Equation: (x−5)² + (y−5)² = 50. Check A(0,0): 25+25=50 ✓ B(10,0): 25+25=50 ✓ C(10,10): 25+25=50 ✓
    • (d) Radius from (5,5) to A(0,0): gradient = 1. Tangent gradient = −1. y = −x ⇒ x + y = 0.