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In the xy-plane, a circle has endpoints of a diameter at \((3, -1)\) and \((-5, 7)\). The equation of the...

GMAT Geometry & Trigonometry : (Geo_Trig) Questions

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Geometry & Trigonometry
Circles
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In the xy-plane, a circle has endpoints of a diameter at \((3, -1)\) and \((-5, 7)\). The equation of the circle can be written in the form \(\mathrm{x}^2 + \mathrm{y}^2 + \mathrm{ax} + \mathrm{by} + \mathrm{c} = 0\), where \(\mathrm{a}\), \(\mathrm{b}\), and \(\mathrm{c}\) are constants. What is the value of \(\mathrm{c}\)?

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Solution

1. TRANSLATE the problem information

  • Given information:
    • Endpoints of a diameter: (3, -1) and (-5, 7)
    • Need to find c in the general form: \(\mathrm{x^2 + y^2 + ax + by + c = 0}\)
  • What this tells us: We have two points that lie on opposite sides of the circle, connected by a line through the center.

2. INFER the approach

  • Since we have diameter endpoints, the center is the midpoint between them
  • The radius is half the distance between the endpoints
  • We'll use standard form first, then expand to general form to identify c

3. SIMPLIFY to find the center

Using the midpoint formula:
Center = \(\mathrm{((3 + (-5))/2, (-1 + 7)/2) = (-1, 3)}\)

4. SIMPLIFY to find the radius

Distance between endpoints = \(\mathrm{\sqrt{[(3-(-5))^2 + (-1-7)^2]} = \sqrt{[64 + 64]} = \sqrt{128} = 8\sqrt{2}}\)
Radius = \(\mathrm{(8\sqrt{2})/2 = 4\sqrt{2}}\)
Therefore: \(\mathrm{r^2 = 32}\)

5. INFER the standard form equation

With center (-1, 3) and \(\mathrm{r^2 = 32}\):
\(\mathrm{(x - (-1))^2 + (y - 3)^2 = 32}\)
\(\mathrm{(x + 1)^2 + (y - 3)^2 = 32}\)

6. SIMPLIFY by expanding to general form

\(\mathrm{(x + 1)^2 + (y - 3)^2 = 32}\)
\(\mathrm{x^2 + 2x + 1 + y^2 - 6y + 9 = 32}\)
\(\mathrm{x^2 + y^2 + 2x - 6y + 10 = 32}\)
\(\mathrm{x^2 + y^2 + 2x - 6y - 22 = 0}\)

Answer: \(\mathrm{c = -22}\)


Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students recognize they need to find the center and radius but incorrectly assume one of the given points IS the center rather than understanding that the center is the midpoint of the diameter endpoints.

This misconception leads them to use (3, -1) or (-5, 7) as the center, then calculate an incorrect radius using the distance to the other point. When they expand their incorrect standard form equation, they get a completely wrong value for c, leading to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly find the center (-1, 3) and radius, but make algebraic errors when expanding \(\mathrm{(x + 1)^2 + (y - 3)^2 = 32}\) to general form.

Common mistakes include sign errors (writing -2x instead of +2x), forgetting to subtract 32 from both sides, or arithmetic errors when combining constants. These errors produce incorrect values for the coefficients, making c wrong despite using the correct approach.

The Bottom Line:

This problem tests whether students understand the geometric relationship between diameter endpoints and the circle's center, combined with careful algebraic manipulation. The key insight is recognizing that diameter endpoints give you the center through the midpoint formula, not that either endpoint IS the center.

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