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A circle in the xy-plane has its center at \((-2, 3)\) and passes through the point \((1, 7)\). An equation...

GMAT Geometry & Trigonometry : (Geo_Trig) Questions

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Geometry & Trigonometry
Circles
HARD
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A circle in the xy-plane has its center at \((-2, 3)\) and passes through the point \((1, 7)\). An equation of this 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:
    • Circle center: \(\mathrm{(-2, 3)}\)
    • Point on circle: \(\mathrm{(1, 7)}\)
    • Need equation in form: \(\mathrm{x^2 + y^2 + ax + by + c = 0}\)
  • What this tells us: We need to find the radius first, then write the circle equation

2. INFER the approach

  • To write any circle equation, we need the center and radius
  • We have the center, but need to calculate the radius using the given point
  • Strategy: Use distance formula → standard form → expand to general form

3. Find the radius using distance formula

  • Distance from center \(\mathrm{(-2, 3)}\) to point \(\mathrm{(1, 7)}\):

\(\mathrm{r = \sqrt{[(1-(-2))^2 + (7-3)^2]}}\)
\(\mathrm{r = \sqrt{[3^2 + 4^2]}}\)
\(\mathrm{r = \sqrt{[9 + 16]}}\)
\(\mathrm{r = \sqrt{25}}\)
\(\mathrm{r = 5}\)


4. Write the standard form equation

  • Using \(\mathrm{(x-h)^2 + (y-k)^2 = r^2}\) with center \(\mathrm{(-2, 3)}\) and radius \(\mathrm{5}\):

\(\mathrm{(x-(-2))^2 + (y-3)^2 = 5^2}\)
\(\mathrm{(x+2)^2 + (y-3)^2 = 25}\)


5. SIMPLIFY by expanding to general form

  • Expand each squared term:
    • \(\mathrm{(x+2)^2 = x^2 + 4x + 4}\)
    • \(\mathrm{(y-3)^2 = y^2 - 6y + 9}\)
  • Substitute and simplify:

\(\mathrm{x^2 + 4x + 4 + y^2 - 6y + 9 = 25}\)
\(\mathrm{x^2 + y^2 + 4x - 6y + 13 = 25}\)
\(\mathrm{x^2 + y^2 + 4x - 6y - 12 = 0}\)


6. INFER the final answer

  • Comparing \(\mathrm{x^2 + y^2 + 4x - 6y - 12 = 0}\) with \(\mathrm{x^2 + y^2 + ax + by + c = 0}\):

\(\mathrm{a = 4, b = -6, c = -12}\)

Answer: c = -12



Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make sign errors when expanding \(\mathrm{(x+2)^2}\) and \(\mathrm{(y-3)^2}\), particularly with the \(\mathrm{(y-3)^2}\) term where they might write \(\mathrm{+6y}\) instead of \(\mathrm{-6y}\), or make errors when combining like terms.

This leads to incorrect values for the coefficients, causing them to calculate the wrong value for c.

Second Most Common Error:

Poor TRANSLATE reasoning: Students might confuse which point is the center versus which point is on the circle, leading them to use the wrong coordinates in the distance formula.

This results in an incorrect radius calculation, which cascades through the entire solution and produces the wrong final equation.

The Bottom Line:

This problem requires careful algebraic manipulation through multiple steps. Success depends on methodical expansion and attention to signs, especially when dealing with negative coordinates in the center.

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