Hint
Complete both squares. The radius direction at P supplies a normal vector to the tangent.
Worked solution
(a)
Completing squares gives \(\displaystyle (x-3)^{2}+(y+2)^{2}=25\). Hence \(\displaystyle C=(3,-2)\) and \(\displaystyle r=5\).
Answer: Centre \((3,-2)\); radius \(5\).
(b)
The displacement CP is (3,4), with squared length 25, so P is on the circle. The tangent is \(\displaystyle 3(x-6)+4(y-2)=0\), or \(\displaystyle 3x+4y=26\).
Answer: \(3x+4y=26\).
(c)
The intercepts are \(\displaystyle (\frac{26}{3}, 0)\) and \(\displaystyle (0, \frac{13}{2})\). Their distance is \(\displaystyle \sqrt{(\frac{26}{3})^{2}+ (\frac{13}{2})^{2}}= \sqrt{\frac{4225}{36}}= \frac{65}{6}\).
Answer: \(\displaystyle \frac{65}{6}\).
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: Complete both squares with correct compensation.
- 1 mark: Give centre (3,−2).
- 1 mark: Give radius 5.
Alternative methods: Coefficient comparison with the standard circle form is accepted.
Common error: Giving radius 25 or reversing a centre-coordinate sign.
(b) · 3 marks
- 1 mark: Verify P on the circle using squared distance or substitution.
- 1 mark: Use a tangent perpendicular to CP through P.
- 1 mark: Obtain \(\displaystyle 3x+4y=26\).
Alternative methods: A negative-reciprocal gradient method is accepted.
Common error: Using the radius direction as the tangent direction.
(c) · 3 marks
- 1 mark: Find both intercept coordinates.
- 1 mark: Use the distance formula between the intercepts.
- 1 mark: Obtain exact length \(\displaystyle \frac{65}{6}\).
Alternative methods: Equivalent exact arithmetic is accepted.
Common error: Adding the two intercept lengths instead of using Pythagoras.
Total: 9 marks.