In the xy-plane, a circle has center \((3, 2)\).The circle intersects the y-axis at the points \((0, -2)\) and \((0,...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
- In the xy-plane, a circle has center \((3, 2)\).
- The circle intersects the y-axis at the points \((0, -2)\) and \((0, 6)\).
- An equation of the circle is \((\mathrm{x} - 3)^2 + (\mathrm{y} - 2)^2 = \mathrm{r}^2\), where \(\mathrm{r}\) is a positive constant.
- What is the value of \(\mathrm{r}\)?
Enter your answer as an integer.
1. TRANSLATE the problem information
- Given information:
- Circle center: \((3, 2)\)
- Y-axis intersection points: \((0, -2)\) and \((0, 6)\)
- Circle equation form: \((\mathrm{x} - 3)^2 + (\mathrm{y} - 2)^2 = \mathrm{r}^2\)
- What this tells us: Any point on the circle must satisfy this equation
2. INFER the solution strategy
- Key insight: Since \((0, -2)\) and \((0, 6)\) are on the circle, they must satisfy the equation
- Strategy: Substitute either intersection point into the equation to solve for r
3. SIMPLIFY by substituting point (0, 6)
- \((0 - 3)^2 + (6 - 2)^2 = \mathrm{r}^2\)
- \((-3)^2 + (4)^2 = \mathrm{r}^2\)
- \(9 + 16 = \mathrm{r}^2\)
- \(25 = \mathrm{r}^2\)
- \(\mathrm{r} = 5\) (taking positive square root since radius must be positive)
4. APPLY CONSTRAINTS to verify the solution
- Check with second point \((0, -2)\): \((0 - 3)^2 + (-2 - 2)^2 = 9 + 16 = 25\) ✓
- Radius must be positive: \(\mathrm{r} = 5\) ✓
Answer: 5
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize that intersection points must satisfy the circle equation. They might try to find the radius using only the center coordinates or attempt to use formulas they don't fully understand (like trying to apply the distance between intersection points as the diameter without proper reasoning). This leads to confusion and guessing.
Second Most Common Error:
Poor SIMPLIFY execution: Students make arithmetic errors when squaring negative numbers, particularly with \((-3)^2\) or \((-4)^2\), or they add incorrectly. For example, they might calculate \((-3)^2\) as \(-9\) instead of 9, leading to \(\mathrm{r}^2 = 7\) and \(\mathrm{r} \approx 2.6\). Since this is a fill-in-the-blank question, this causes them to get stuck with a non-integer answer.
The Bottom Line:
This problem tests whether students understand the fundamental definition of a circle—that every point on the circle is exactly r units from the center. Many students know the circle equation but fail to connect it with the practical constraint that given points must satisfy this equation.