Hint
The centre is the diameter midpoint. The point opposite T is its reflection through that centre.
Worked solution
(a)
The centre is C=(2,3). The diameter squared is 8²+4²=80, so r²=80/4=20. The equation follows.
Answer: \((x-2)^2+(y-3)^2=20\).
(b)
CT=(2,−4), and 2²+(−4)²=20, so T is on the circle. The tangent is 2(x−4)−4(y+1)=0, which simplifies to x−2y=6.
Answer: \(x-2y=6\).
(c)
The opposite point is U=2C−T=(0,7). Its radius is parallel to CT, so its tangent is parallel to the first. Substituting U into x−2y=c gives c=−14.
Answer: \(x-2y=-14\).
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: Find centre (2,3).
- 1 mark: Obtain squared radius 20 from the diameter.
- 1 mark: Give the complete circle equation.
Alternative methods: Using the distance from C to either diameter endpoint is accepted.
Common error: Using 80 as r².
(b) · 3 marks
- 1 mark: Verify that T is on the circle.
- 1 mark: Use CT as a normal to the tangent through T.
- 1 mark: Obtain x−2y=6.
Alternative methods: A correct perpendicular-gradient method is accepted.
Common error: Ignoring the negative y-displacement of CT.
(c) · 3 marks
- 1 mark: Find the opposite point (0,7), or an equivalent symmetric offset.
- 1 mark: Use a parallel tangent direction at that opposite point.
- 1 mark: Obtain x−2y=−14.
Alternative methods: The centre-to-line distance condition for the two parallel tangents is accepted.
Common error: Giving a parallel line through the centre, which is not a tangent.
Total: 9 marks.