Question:In the xy-plane, a circle has equation x^2 + y^2 - 4x + 6y = 23. Which of the following...

1. TRANSLATE the problem information

• Given information:
  • Circle equation: \(\mathrm{x^2 + y^2 - 4x + 6y = 23}\)
  • Five points to test: (2,-3), (4,-6), (-1,-5), (8,-3), (0,-1)
  • Need to find which point is NOT in the interior
• What this tells us: We need to determine the circle's center and radius, then test each point's position relative to the circle.

2. INFER the approach

• To determine if points are interior, we need the center and radius from the equation
• The equation is in general form, so we need to convert to standard form \(\mathrm{(x-h)^2 + (y-k)^2 = r^2}\)
• Once we have center and radius, we can check distances from center to each point

3. SIMPLIFY the circle equation by completing the square

• Rearrange: \(\mathrm{x^2 - 4x + y^2 + 6y = 23}\)
• For x-terms: \(\mathrm{x^2 - 4x}\)
  • Complete the square: \(\mathrm{(x - 2)^2 = x^2 - 4x + 4}\)
  • So \(\mathrm{x^2 - 4x = (x - 2)^2 - 4}\)
• For y-terms: \(\mathrm{y^2 + 6y}\)
  • Complete the square: \(\mathrm{(y + 3)^2 = y^2 + 6y + 9}\)
  • So \(\mathrm{y^2 + 6y = (y + 3)^2 - 9}\)
• Substitute back: \(\mathrm{(x - 2)^2 - 4 + (y + 3)^2 - 9 = 23}\)
• SIMPLIFY: \(\mathrm{(x - 2)^2 + (y + 3)^2 = 36}\)

4. INFER the circle properties

• From \(\mathrm{(x - 2)^2 + (y + 3)^2 = 36}\):
  • Center: (2, -3)
  • Radius: \(\mathrm{\sqrt{36} = 6}\)

5. INFER the testing strategy

• A point is interior if its distance from center < radius
• We can use distance squared to avoid square roots: \(\mathrm{d^2 \lt r^2}\) means interior
• Since \(\mathrm{r^2 = 36}\), we need \(\mathrm{d^2 \lt 36}\) for interior points

6. SIMPLIFY distance calculations for each point

• Point (A) (2, -3): \(\mathrm{d^2 = (2-2)^2 + (-3-(-3))^2 = 0^2 + 0^2 = 0 \lt 36}\) ✓ Interior
• Point (B) (4, -6): \(\mathrm{d^2 = (4-2)^2 + (-6-(-3))^2 = 2^2 + (-3)^2 = 4 + 9 = 13 \lt 36}\) ✓ Interior
• Point (C) (-1, -5): \(\mathrm{d^2 = (-1-2)^2 + (-5-(-3))^2 = (-3)^2 + (-2)^2 = 9 + 4 = 13 \lt 36}\) ✓ Interior
• Point (D) (8, -3): \(\mathrm{d^2 = (8-2)^2 + (-3-(-3))^2 = 6^2 + 0^2 = 36 = 36}\) ⚠️ On boundary
• Point (E) (0, -1): \(\mathrm{d^2 = (0-2)^2 + (-1-(-3))^2 = (-2)^2 + 2^2 = 4 + 4 = 8 \lt 36}\) ✓ Interior

7. APPLY CONSTRAINTS to identify the answer

• Point D has \(\mathrm{d^2 = 36 = r^2}\), so it lies exactly on the circle boundary
• Points on the boundary are not considered interior points
• All other points have \(\mathrm{d^2 \lt 36}\), making them interior

Answer: D

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students struggle with completing the square, particularly managing the constant terms correctly. They might forget to subtract the added constants (4 and 9) or make sign errors when moving terms around.

This leads to incorrect center/radius identification, causing them to misclassify all the points. This causes confusion and guessing.

Second Most Common Error:

Missing conceptual knowledge about circle interior: Students may not realize that points exactly on the circle boundary (where \(\mathrm{d = r}\)) are not considered interior points. They might think "on the circle" means "inside the circle."

This may lead them to incorrectly eliminate choice D as a valid answer and select a point that is genuinely interior, such as Choice A (2, -3) since it's the center.

The Bottom Line:

This problem requires solid algebraic manipulation skills for completing the square AND precise understanding of geometric terminology. The distinction between "interior," "on the boundary," and "exterior" is crucial for the correct answer.