The graph of 9x^2 + 18x + 9y^2 + 6y = 890 in the xy-plane is a circle. What is...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
The graph of \(\mathrm{9x^2 + 18x + 9y^2 + 6y = 890}\) in the xy-plane is a circle. What is the length of the circle's radius?
- 5
- 10
- 15
- 30
- 100
1. TRANSLATE the problem information
- Given: \(9\mathrm{x}^2 + 18\mathrm{x} + 9\mathrm{y}^2 + 6\mathrm{y} = 890\) represents a circle
- Find: The radius length
2. INFER the solution approach
- To find the radius, we need the equation in standard form: \((\mathrm{x} - \mathrm{h})^2 + (\mathrm{y} - \mathrm{k})^2 = \mathrm{r}^2\)
- This requires completing the square for both x and y terms
- Start by simplifying coefficients to make completing the square easier
3. SIMPLIFY by dividing through by the common factor
- Divide both sides by 9:
- \(\mathrm{x}^2 + 2\mathrm{x} + \mathrm{y}^2 + \frac{2}{3}\mathrm{y} = \frac{890}{9}\)
4. SIMPLIFY by completing the square for x
- \(\mathrm{x}^2 + 2\mathrm{x} = (\mathrm{x} + 1)^2 - 1\)
- Take half of the coefficient of x (which is 2), square it: \((\frac{2}{2})^2 = 1\)
- Add and subtract this inside the expression
5. SIMPLIFY by completing the square for y
- \(\mathrm{y}^2 + \frac{2}{3}\mathrm{y} = (\mathrm{y} + \frac{1}{3})^2 - (\frac{1}{3})^2\)
- Take half of \(\frac{2}{3}\): \(\frac{2}{3} \div 2 = \frac{1}{3}\)
- Square it: \((\frac{1}{3})^2 = \frac{1}{9}\)
- So \(\mathrm{y}^2 + \frac{2}{3}\mathrm{y} = (\mathrm{y} + \frac{1}{3})^2 - \frac{1}{9}\)
6. SIMPLIFY by substituting and combining constants
- \((\mathrm{x} + 1)^2 - 1 + (\mathrm{y} + \frac{1}{3})^2 - \frac{1}{9} = \frac{890}{9}\)
- \((\mathrm{x} + 1)^2 + (\mathrm{y} + \frac{1}{3})^2 = \frac{890}{9} + 1 + \frac{1}{9}\)
- Convert 1 to ninths: \(1 = \frac{9}{9}\)
- \((\mathrm{x} + 1)^2 + (\mathrm{y} + \frac{1}{3})^2 = \frac{890}{9} + \frac{9}{9} + \frac{1}{9} = \frac{900}{9} = 100\)
7. INFER the radius from standard form
- We now have \((\mathrm{x} + 1)^2 + (\mathrm{y} + \frac{1}{3})^2 = 100\)
- Comparing to \((\mathrm{x} - \mathrm{h})^2 + (\mathrm{y} - \mathrm{k})^2 = \mathrm{r}^2\), we see \(\mathrm{r}^2 = 100\)
- Therefore \(\mathrm{r} = \sqrt{100} = 10\)
Answer: B) 10
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Students make arithmetic errors when combining the constant terms at the end. They might incorrectly calculate \(\frac{890}{9} + 1 + \frac{1}{9}\), especially struggling with converting 1 to the fraction \(\frac{9}{9}\). This leads to getting \(\mathrm{r}^2\) equal to something other than 100, resulting in an incorrect radius calculation. This may lead them to select Choice A (5) or Choice C (15) depending on their specific arithmetic mistake.
Second Most Common Error:
Inadequate INFER reasoning: Students attempt to find the radius without converting to standard form, perhaps trying to use the discriminant or other inappropriate methods. They might also forget that after finding \(\mathrm{r}^2 = 100\), they need to take the square root to get the actual radius. This leads to confusion and guessing among the answer choices.
The Bottom Line:
This problem requires systematic algebraic manipulation through completing the square - a multi-step process where small errors compound. Success depends on careful arithmetic with fractions and recognizing the connection between standard form and radius identification.