Hint
Substitute the line into the circle. Match each y-value with its own x-value before finding the distance.
Worked solution
(a)
Expanding gives \(\displaystyle 2x^{2}+4x+4=20\). Rearrange and divide by 2 to obtain \(\displaystyle x^{2}+2x-8=0\).
Answer: \(x^2+(x+2)^2=20\), hence \(x^2+2x-8=0\).
(b)
\(\displaystyle (x+4)(x-2)=0\), so \(\displaystyle x=-4\) or \(\displaystyle x=2\). Using \(\displaystyle y=x+2\) gives (−4,−2) and (2,4). Each satisfies the circle equation.
Answer: \((-4,-2)\) and \((2,4)\), in either order.
(c)
\(\displaystyle AB=\sqrt{(2+4)^{2}+(4+2)^{2}}=\sqrt{72}=6\sqrt{2}\).
Answer: \(6\sqrt2\).
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) · 2 marks
- 1 mark: Substitute \(\displaystyle y=x+2\) into the circle correctly.
- 1 mark: Expand and simplify to the required equation.
Alternative methods: Eliminating x instead of y is valid if the requested x-equation is recovered.
Common error: Expanding \(\displaystyle (x+2)^{2}\) as \(\displaystyle x^{2}+4\).
(b) · 3 marks
- 1 mark: Obtain both x-values −4 and 2.
- 1 mark: Calculate the corresponding y-values.
- 1 mark: State both correctly paired points.
Alternative methods: The quadratic formula and substitution are accepted.
Common error: Matching −4 with 4 or 2 with −2.
(c) · 2 marks
- 1 mark: Apply the distance formula to the two points.
- 1 mark: Simplify to \(\displaystyle 6\sqrt{2}\); method credit may follow through two distinct points from part (b).
Alternative methods: Pythagoras using horizontal and vertical differences 6 and 6 is accepted.
Common error: Adding coordinate differences instead of their squares.
Total: 7 marks.