Question:In the xy-plane, the equation x^2 + y^2 + 4x - 6y + k = 0 represents a circle, where...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
In the xy-plane, the equation \(\mathrm{x^2 + y^2 + 4x - 6y + k = 0}\) represents a circle, where \(\mathrm{k}\) is a real constant. The circle has radius \(\mathrm{7}\). What is the value of \(\mathrm{k}\)?
Enter your answer as an integer.
No answer choices — enter an integer.
1. TRANSLATE the problem information
- Given information:
- Equation: \(\mathrm{x^2 + y^2 + 4x - 6y + k = 0}\) represents a circle
- The circle has radius 7
- Need to find the value of k
2. INFER the solution approach
- To work with circle problems, we need the standard form: \(\mathrm{(x - h)^2 + (y - k)^2 = r^2}\)
- Our equation is in general form, so we must complete the square for both x and y terms
- Once in standard form, we can use the given radius to find k
3. SIMPLIFY by completing the square for x terms
- Start with: \(\mathrm{x^2 + 4x}\)
- To complete the square: take half of the coefficient of x, then square it
- Half of 4 is 2, and \(\mathrm{2^2 = 4}\)
- So: \(\mathrm{x^2 + 4x = (x + 2)^2 - 4}\)
4. SIMPLIFY by completing the square for y terms
- Start with: \(\mathrm{y^2 - 6y}\)
- Half of -6 is -3, and \(\mathrm{(-3)^2 = 9}\)
- So: \(\mathrm{y^2 - 6y = (y - 3)^2 - 9}\)
5. SIMPLIFY by substituting back into the original equation
- Replace the x and y terms in the original equation:
- \(\mathrm{(x + 2)^2 - 4 + (y - 3)^2 - 9 + k = 0}\)
- Rearrange: \(\mathrm{(x + 2)^2 + (y - 3)^2 = 4 + 9 - k}\)
- Simplify: \(\mathrm{(x + 2)^2 + (y - 3)^2 = 13 - k}\)
6. INFER the final relationship and solve
- Our equation is now in standard form with \(\mathrm{r^2 = 13 - k}\)
- Since the radius is 7, we have \(\mathrm{r^2 = 49}\)
- Set up the equation: \(\mathrm{13 - k = 49}\)
- Solve: \(\mathrm{k = 13 - 49 = -36}\)
Answer: -36
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Students make sign errors when completing the square, especially with the y terms where the coefficient is negative (-6y). They might forget that \(\mathrm{(y - 3)^2 - 9}\) expands to \(\mathrm{y^2 - 6y + 9 - 9 = y^2 - 6y}\), not \(\mathrm{y^2 + 6y}\).
This algebraic error propagates through the solution, leading to an incorrect final equation and wrong value for k. This leads to confusion and potentially guessing among answer options.
Second Most Common Error:
Missing conceptual knowledge: Students may not remember the completing the square technique or confuse which constant term to add/subtract. Some students might attempt to force the given equation directly into standard form without properly completing the square.
This causes them to get stuck early in the problem and abandon systematic solution, leading to guessing.
The Bottom Line:
This problem tests whether students can execute the multi-step algebraic process of completing the square accurately while keeping track of all terms. The key challenge is maintaining precision through several algebraic manipulations.