Hint
After substitution, the two square terms are equal. Keep each x- and y-coordinate on the same square-root branch.
Worked solution
(a)
Substituting gives 2(x−1)²=25. Thus x=1±5√2/2. Since y=x−3, the corresponding outputs are −2±5√2/2, with matched signs.
Answer: \(A=(1+5\sqrt2/2,-2+5\sqrt2/2)\), \(B=(1-5\sqrt2/2,-2-5\sqrt2/2)\), in either order.
(b)
Averaging cancels the surds and gives midpoint (1,−2), the circle centre. The chord is therefore a diameter with length 2r=10. Directly, the coordinate differences are both 5√2, giving the same length.
Answer: Midpoint \((1,-2)\); \(AB=10\), a diameter.
(c)
The chord line has gradient 1. Its bisector has gradient −1 and passes through (1,−2), giving y+2=−(x−1), or x+y=−1.
Answer: \(x+y=-1\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Obtain 2(x−1)²=25 or an equivalent quadratic.
- 1 mark: Find both exact x-values.
- 1 mark: Give both correctly paired coordinates.
Alternative methods: The quadratic formula gives equivalent exact surds.
Common error: Mixing the signs of the x- and y-coordinates.
(b) · 3 marks
- 1 mark: Find midpoint (1,−2).
- 1 mark: Recognise a diameter through the centre or use the correct distance calculation.
- 1 mark: Obtain AB=10.
Alternative methods: A complete distance-formula calculation with the centre identified is accepted.
Common error: Using the radius 5 as the whole chord length.
(c) · 2 marks
- 1 mark: Use gradient −1 through the midpoint.
- 1 mark: Obtain x+y=−1.
Alternative methods: An equal-distance derivation is accepted.
Common error: Giving the original chord line instead of its perpendicular bisector.
Total: 8 marks.