In the xy-plane, a circle has center O with coordinates \(\mathrm{(c,d)}\). Points P and Q are on the circle, and...

1. TRANSLATE the coordinate information

• Given information:
  • Circle center O is at \(\mathrm{(c,d)}\)
  • Chord PQ has length 16
  • Midpoint M of chord PQ is at \(\mathrm{(c, d-6)}\)
• What this tells us: The center and chord midpoint have the same x-coordinate, so they form a vertical line

2. INFER the geometric relationship

• Key insight: A line from the center of a circle to the midpoint of any chord is always perpendicular to that chord
• This creates a right triangle where:
  • OM is one leg (center to midpoint)
  • PM is the other leg (midpoint to chord endpoint)
  • OP is the hypotenuse (which equals the radius)

3. SIMPLIFY the distance calculation

• Distance from \(\mathrm{O(c,d)}\) to \(\mathrm{M(c, d-6)}\):
  \(\mathrm{OM = \sqrt{(c-c)^2 + (d-(d-6))^2}}\)
  \(\mathrm{= \sqrt{0^2 + 6^2}}\)
  \(\mathrm{= 6}\)

4. INFER the chord measurement

• Since M is the midpoint of chord PQ with length 16:
  \(\mathrm{PM = PQ \div 2}\)
  \(\mathrm{= 16 \div 2}\)
  \(\mathrm{= 8}\)

5. SIMPLIFY using Pythagorean theorem

• In right triangle OMP:
  \(\mathrm{r^2 = OM^2 + PM^2}\)
  \(\mathrm{r^2 = 6^2 + 8^2}\)
  \(\mathrm{r^2 = 36 + 64 = 100}\)
  \(\mathrm{r = 10}\)

6. TRANSLATE to area formula

• Area:
  \(\mathrm{Area = \pi r^2}\)
  \(\mathrm{= \pi(10)^2}\)
  \(\mathrm{= 100\pi}\)

Answer: C) 100π

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students fail to recognize the perpendicular relationship between the line from center to chord midpoint and the chord itself. Instead, they might try to use the distance formula directly from center to chord endpoints or attempt complex coordinate geometry approaches. Without recognizing this creates a right triangle, they get stuck in complicated calculations and end up guessing.

Second Most Common Error:

Incomplete SIMPLIFY execution: Students correctly set up the Pythagorean theorem but make arithmetic errors in the calculation. For example, calculating \(\mathrm{6^2 + 8^2 = 36 + 64 = 90}\) instead of 100, leading to \(\mathrm{r^2 = 90}\) and area = \(\mathrm{90\pi}\). Since \(\mathrm{90\pi}\) isn't among the choices, this leads to confusion and random selection.

The Bottom Line:

This problem tests whether students can recognize fundamental circle properties (perpendicular from center to chord) and apply them within a coordinate geometry context. The key breakthrough is seeing that the given coordinates create a clean right triangle setup.