Solutions — Graphs of Relations — Circles and Parabolas

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1. Q1 — Centre and radius

   (a) Centre (0,0), radius 6   (b) Centre (1,4), radius 3   (c) Centre (−3,2), radius 5   (d) Centre (0,−5), radius 10

2. Q2 — Writing circle equations

   (a) x² + y² = 64   (b) (x−3)² + (y+1)² = 25   (c) (x+2)² + (y−4)² = 7

3. Q3 — Horizontal parabola y² = x

   (a) Vertex (0,0); opens to the right   (b) x=9: y²=9, y=±3   (c) y²=x has two y-values for each x>0 (fails vertical line test). y=√x restricts to the upper branch only (y≥0), giving exactly one output per input → it is a function.   (d) Upper half of y²=4x: y = 2√x (domain x≥0, range y≥0)

4. Q4 — y = √(x−2) + 1

   Domain: x−2≥0 → x ≥ 2. Range: √(x−2)≥0, so y≥1. Range: y ≥ 1. Endpoint (start of graph): (2, 1).

5. Q5 — Complete the square

   (a) (x²−6x+9)+(y²+2y+1)=6+9+1 → (x−3)²+(y+1)²=16. Centre (3,−1), radius 4.   (b) (x²+4x+4)+(y²−8y+16)=−11+4+16 → (x+2)²+(y−4)²=9. Centre (−2,4), radius 3.

6. Q6 — Inside, on, or outside circle (x−1)²+(y+2)²=25

   (a) (4,2): (4−1)²+(2+2)²=9+16=25. On the circle.   (b) (−3,−2): (−4)²+(0)²=16<25. Inside the circle.   (c) (0,0): (0−1)²+(0+2)²=1+4=5<25. Inside the circle.

7. Q7 — Intercepts and intersections

   (a) y=0: x²=25 → x=±5. x-intercepts: (−5,0) and (5,0).   (b) x=0: (0−3)²+y²=16 → y²=7 → y=±√7. y-intercepts: (0,√7) and (0,−√7).   (c) Substitute y=x+3 into x²+y²=4: x²+(x+3)²=4 → 2x²+6x+9=4 → 2x²+6x+5=0. Δ=36−40=−4<0. No real solutions — the line and circle do not intersect.

8. Q8 — Square root functions

   (a) y=√x is the upper half; y=−√x is the lower half. Together they form the horizontal parabola y²=x.   (b) y=2√(x−3)+1: Start with y=√x. Shift right 3 (x−3), stretch vertically by factor 2, shift up 1. Domain: x≥3. Range: y≥1.

9. Q9 — Circle from conditions

   (a) r=√((5−2)²+(3−(−1))²)=√(9+16)=√25=5. Equation: (x−2)²+(y+1)²=25.   (b) Centre = midpoint = ((1+7)/2, (2+8)/2) = (4,5). r=√((7−4)²+(8−5)²)=√(9+9)=3√2. Equation: (x−4)²+(y−5)²=18.   (c) Circle centre (3,0), tangent to y-axis means radius=3 (horizontal distance to y-axis). Equation: (x−3)²+y²=9. Point of tangency: (0,0).

10. Q10 — Circular tunnel problem

    (a) Centre (0,4), radius 5. The tunnel’s cross-section is centred 4 m above the road.   (b) At y=0: x²+(0−4)²=25 → x²=9 → x=±3. Width = 6 m.   (c) Truck is 2.8 m wide (±1.4 from centre) and 4 m tall. Check point (1.4, 4): (1.4)²+(4−4)²=1.96<25 ✓. So the top of the truck at height 4 m is well inside. Also check point (1.4, 0): above calculated. The truck’s roof is at y=4 m, and at x=±1.4, the circle height is: (1.4)²+(y−4)²=25 → (y−4)²=23.04 → y=4+√23.04≈8.8 m > 4 m. The truck fits.