Try this
“The equation of the curve is ” demands that two things both be true:
- Purity: the points of the curve all satisfy the equation. Put the other way round, every point given by a solution of the equation satisfies the condition: no extra points have slipped in;
- Completeness: every point that satisfies the condition is on the curve: not a single one is missed.
In the widget, the small dots are the locus points the mechanism “really” produces, and the dashed curve is an equation somebody derived. is your test point, and two readouts check “does it satisfy the condition?” and “does it satisfy the equation?” separately.
- Choose “Chord midpoints” and drag the slider from to . Which part of the dashed circle do the dots cover? Beyond what does the line stop cutting the circle?
- Press “” and “ itself”. What are the verdicts?
- Choose “Right-angle vertex” and drag the slider back and forth. What happens at ? Then send to .
- Choose “Equidistant from the axes” and drag the slider through its range. How many lines do the dots draw? How many dashed lines are there? Send to .
- Of the four verdicts (genuine, extra, missing, unrelated), which one shows the equation is not pure? Which one shows it is not complete?
Predict first
Lines through cut chords from the circle , and the chord midpoints lie on the circle . How much of that circle do they occupy?
Subtracting from gives the intersection points of the two circles: , . Seen from the center , they are in the directions and , and the arc between them is of the circle.
Only the points of this arc (the part inside the given circle, without the two tangency points at its ends) are real chord midpoints. For every other point, included, the line through it and does not cut the circle at all.
Why
The standard procedure for a locus problem is: set up coordinates and a moving point → write the condition → turn it into an equation → check. The last step is exactly the check of purity and completeness. Here is a complete run.
Example: lines through P₀(4, 0) cut the circle x² + y² = 4; find the locus of the chord midpoint M
Step 1 / 5- Find the condition
The line joining the midpoint of a chord to the center is perpendicular to the chord: . So , and lies on the circle with diameter .
Notice that the final range was dug out of the condition, not out of the equation: the equation itself knows nothing about the given circle. That is exactly why purity has to be checked separately.
Practice
, . The locus of points with is the circle with some points removed. How many points are removed?
How many lines make up the locus of points whose distance to the -axis is times their distance to the -axis?
Chords of the circle are drawn through the interior point . The locus of their midpoints is a complete circle. What is its radius?
Lines through cut chords from the circle . The locus of the chord midpoints is the arc of with . Find .