In the xy-plane, a circle has a diameter with endpoints at \((-1, 4)\) and \((5, -2)\). Which equation represents this...

1. TRANSLATE the problem information

• Given information:
  • Circle has diameter with endpoints (-1, 4) and (5, -2)
  • Need to find the equation representing this circle

• What this tells us: We need to find the center and radius to write the circle equation in standard form \((\mathrm{x} - \mathrm{h})^2 + (\mathrm{y} - \mathrm{k})^2 = \mathrm{r}^2\)

2. INFER the approach

• Since we have diameter endpoints, the center must be the midpoint between them
• The radius will be half the distance between the endpoints (since radius = diameter/2)
• This gives us everything needed for the standard form equation

3. SIMPLIFY to find the center

• Using midpoint formula with (-1, 4) and (5, -2):
• Center = \(\left(\frac{-1 + 5}{2}, \frac{4 + (-2)}{2}\right) = \left(\frac{4}{2}, \frac{2}{2}\right) = (2, 1)\)

4. SIMPLIFY to find the radius

• First find distance between endpoints:
• \(\mathrm{d} = \sqrt{(5 - (-1))^2 + (-2 - 4)^2}\)
• \(= \sqrt{6^2 + (-6)^2}\)
• \(= \sqrt{36 + 36}\)
• \(= \sqrt{72}\)
• SIMPLIFY \(\sqrt{72}\): \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\)
• Radius = diameter/2 = \(\frac{6\sqrt{2}}{2} = 3\sqrt{2}\)
• Therefore: \(\mathrm{r}^2 = (3\sqrt{2})^2 = 9 \times 2 = 18\)

5. TRANSLATE to standard form

• With center (2, 1) and \(\mathrm{r}^2 = 18\):
• \((\mathrm{x} - 2)^2 + (\mathrm{y} - 1)^2 = 18\)

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that having diameter endpoints means they need to find the midpoint for the center. Instead, they might try using one of the endpoints as the center, leading to incorrect calculations.

This might lead them to select Choice A if they use (-1, 4) as center, or get completely confused and guess.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly find that they need half the distance between endpoints for radius, but make calculation errors. They might use the full distance \((6\sqrt{2})\) instead of the radius \((3\sqrt{2})\), leading to \(\mathrm{r}^2 = 72\) instead of \(\mathrm{r}^2 = 18\).

This may lead them to select Choice D which has \(\mathrm{r}^2 = 36\), or cause calculation confusion.

The Bottom Line:

This problem tests whether students understand the relationship between diameter endpoints and circle properties. The key insight is recognizing that diameter endpoints give you everything needed - just find the midpoint for center and half the distance for radius.