Kenneth ChengMathematical modeling
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A moving point stays a fixed distance from (2,3), and its locus passes through the origin. Find the locus.

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TOPIC 01

2024 JM01

The constant cancels because the circle passes through (0,0).

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

A moving point stays a fixed distance from (2,3), and its locus passes through the origin. Find the locus.

Official paper · jm01-2024 · I.5 · PDF 2

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option Ax2+y2−13=0x^2+y^2-13=0
  2. Option Bx2+y2+4x−6y=0x^2+y^2+4x-6y=0
  3. Option Cx2+y2+4x+6y=0x^2+y^2+4x+6y=0
  4. Option Dx2+y2−4x−6y=0x^2+y^2-4x-6y=0
  5. Option Ex2+y2−4x−6y+13=0x^2+y^2-4x-6y+13=0

Working and explanation

BUILD THE REASONING

Hint 1
The fixed point is the circle centre.
Hint 2
Use its distance from the origin as the radius.
Worked solution

  1. Square the radius.

    r2=22+32=13r^2=2^2+3^2=13

  2. Expand the standard circle equation.

    (x−2)2+(y−3)2=13  ⟺  x2+y2−4x−6y=0(x-2)^2+(y-3)^2=13\iff x^2+y^2-4x-6y=0

D.

Checks and common pitfalls: The constant cancels because the circle passes through (0,0).

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Square the radius.
    r2=22+32=13r^2=2^2+3^2=13
  • Expand the standard circle equation.
    (x−2)2+(y−3)2=13  ⟺  x2+y2−4x−6y=0(x-2)^2+(y-3)^2=13\iff x^2+y^2-4x-6y=0

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