Question:A circle in the xy-plane has a diameter with endpoints at \((1, -1)\) and \((9, 5)\). Which of the following...

1. TRANSLATE the problem information

• Given information:
  • Diameter endpoints: \((1, -1)\) and \((9, 5)\)
  • Need: equation of the circle
• This tells us we need to find the center and radius to write the standard circle equation

2. INFER the approach

• To write a circle equation in standard form \((\mathrm{x} - \mathrm{h})^2 + (\mathrm{y} - \mathrm{k})^2 = \mathrm{r}^2\), we need:
  • Center \((\mathrm{h}, \mathrm{k})\): This is the midpoint of the diameter
  • Radius \(\mathrm{r}\): This is the distance from center to either endpoint
• Start with finding the center, then calculate the radius

3. SIMPLIFY to find the center coordinates

• Use midpoint formula with endpoints \((1, -1)\) and \((9, 5)\):
  • \(\mathrm{h} = \frac{1 + 9}{2} = \frac{10}{2} = 5\)
  • \(\mathrm{k} = \frac{-1 + 5}{2} = \frac{4}{2} = 2\)
• Center: \((5, 2)\)

4. SIMPLIFY to find the radius

• Use distance formula from center \((5, 2)\) to endpoint \((1, -1)\):
  • \(\mathrm{r} = \sqrt{(1-5)^2 + (-1-2)^2}\)
  • \(= \sqrt{(-4)^2 + (-3)^2}\)
  • \(= \sqrt{16 + 9}\)
  • \(= \sqrt{25} = 5\)
• Therefore: \(\mathrm{r}^2 = 25\)

5. TRANSLATE into final equation

• Substitute center \((5, 2)\) and \(\mathrm{r}^2 = 25\) into standard form:
• \((\mathrm{x} - 5)^2 + (\mathrm{y} - 2)^2 = 25\)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Making arithmetic mistakes in the midpoint calculation, especially with negative coordinates.

Students often get confused with the signs when calculating \(\mathrm{k} = \frac{-1 + 5}{2}\), writing it as \(\mathrm{k} = \frac{1 + 5}{2} = 3\) instead of \(\mathrm{k} = 2\). This leads to center \((5, 3)\) instead of \((5, 2)\).

This may lead them to select Choice A: \((\mathrm{x} - 4)^2 + (\mathrm{y} - 3)^2 = 25\) (though they'd also need to make an error in the h-coordinate).

Second Most Common Error:

Conceptual confusion about radius vs diameter: Using the full distance between endpoints as \(\mathrm{r}^2\) instead of recognizing it as the diameter.

Students calculate the distance between endpoints: \(\sqrt{(9-1)^2 + (5-(-1))^2} = \sqrt{64 + 36} = \sqrt{100} = 10\), then mistakenly use this as the radius, giving \(\mathrm{r}^2 = 100\).

This may lead them to select Choice D: \((\mathrm{x} - 5)^2 + (\mathrm{y} - 2)^2 = 100\).

The Bottom Line:

This problem tests your understanding of the relationship between diameter and center/radius, combined with careful coordinate arithmetic. The key insight is recognizing that you need the midpoint (not the endpoints) and half the diameter length (not the full distance).