Question:In the xy-plane, circle P consists of all points that are sqrt(13) units from the point \((-2, 3)\).Circle Q has...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
- In the xy-plane, circle P consists of all points that are \(\sqrt{13}\) units from the point \((-2, 3)\).
- Circle Q has the same center as circle P, and its circumference is twice the circumference of circle P.
- The equation of circle Q in standard form can be written as \((\mathrm{x} + 2)^2 + (\mathrm{y} - 3)^2 = \mathrm{t}\), where \(\mathrm{t}\) is a constant. What is the value of \(\mathrm{t}\)?
Enter your answer as an integer.
1. TRANSLATE the problem information
- Given information:
- Circle P: all points \(\sqrt{13}\) units from (-2, 3)
- Circle Q: same center as P, circumference is twice P's circumference
- Circle Q equation: \(\mathrm{(x + 2)^2 + (y - 3)^2 = t}\)
- What this tells us:
- Circle P has center (-2, 3) and radius \(\sqrt{13}\)
- Circle Q has center (-2, 3) and some larger radius
2. INFER how circumference scaling affects radius
- Since circumference \(\mathrm{C = 2\pi r}\), doubling circumference means doubling radius
- If radius of P = \(\sqrt{13}\), then radius of Q = \(2\sqrt{13}\)
- This is the key insight: circumference and radius scale by the same factor
3. SIMPLIFY to find the constant t
- Standard circle form: \(\mathrm{(x - h)^2 + (y - k)^2 = r^2}\)
- For circle Q: \(\mathrm{(x + 2)^2 + (y - 3)^2}\) = (radius of Q)\(\mathrm{^2}\)
- Substitute: \(\mathrm{(x + 2)^2 + (y - 3)^2 = (2\sqrt{13})^2}\)
- Calculate: \(\mathrm{(2\sqrt{13})^2 = 4 \times 13 = 52}\)
Answer: 52
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students think "twice the circumference" means the constant t should be doubled directly.
They might reason: "If P has constant 13, then Q should have constant \(\mathrm{2 \times 13 = 26}\)." This misses that circumference scaling affects radius, which then gets squared in the standard form. This leads them to answer 26.
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly identify that radius doubles but make arithmetic errors in computing \(\mathrm{(2\sqrt{13})^2}\).
They might calculate \(\mathrm{(2\sqrt{13})^2 = 2 \times 13 = 26}\), forgetting that \(\mathrm{(2\sqrt{13})^2 = 2^2 \times (\sqrt{13})^2 = 4 \times 13}\). This also leads to the incorrect answer 26.
The Bottom Line:
This problem tests understanding of how scaling factors propagate through geometric relationships. Students must recognize that linear scaling (circumference) becomes quadratic scaling (area/constant) due to the \(\mathrm{r^2}\) term in circle equations.