Hint
Set the discriminant of the intersection quadratic to zero. For c=7 the line passes through the centre, which checks the final geometry.
Worked solution
(a)
Substitute y=3x+c: (x+2)²+(3x+c−1)²=10. Expand and collect to get 10x²+[4+6(c−1)]x+4+(c−1)²−10=0, giving the stated result.
Answer: \(10x^2+(6c-2)x+(c-1)^2-6=0\).
(b)
The discriminant is (6c−2)²−40[(c−1)²−6]=4[100−(c−7)²]. Setting it to zero gives c−7=±10, hence c=−3 or 17.
Answer: \(c=-3\) or \(c=17\).
(c)
The quadratic becomes 10x²+40x+30=0, or (x+1)(x+3)=0. The roots are −1 and −3. Substitution into y=3x+7 gives (−1,4) and (−3,−2). Their midpoint is the centre (−2,1).
Answer: \((-1,4)\) and \((-3,-2)\).
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: Substitute the line into the circle correctly.
- 1 mark: Obtain quadratic coefficient 10 and linear coefficient 6c−2.
- 1 mark: Obtain the correct constant and complete equation.
Alternative methods: A consistent equivalent expanded quadratic is accepted.
Common error: Dropping the cross term when expanding the second square.
(b) · 3 marks
- 1 mark: Set a correct discriminant to zero or use the centre-to-line distance condition.
- 1 mark: Obtain (c−7)²=100 or its equivalent.
- 1 mark: Give both values −3 and 17.
Alternative methods: The distance from (−2,1) to 3x−y+c=0 is |c−7|/√10; equating it to √10 is accepted.
Common error: Keeping only one of the two parallel tangents.
(c) · 3 marks
- 1 mark: Obtain the correct quadratic for c=7.
- 1 mark: Find both roots −1 and −3.
- 1 mark: Give the two correctly paired points.
Alternative methods: A parameterisation of the central line and radius distance is accepted.
Common error: Substituting a tangent c-value instead of the specified c=7.
Total: 9 marks.