Question:In the coordinate plane, the endpoints of a diameter of a circle are \(\mathrm{A(-5, 7)}\) and \(\mathrm{B(7, -11)}\). The radius...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
In the coordinate plane, the endpoints of a diameter of a circle are \(\mathrm{A(-5, 7)}\) and \(\mathrm{B(7, -11)}\). The radius of the circle can be written in the form \(\mathrm{3\sqrt{d}}\), where \(\mathrm{d}\) is a positive integer. What is the value of \(\mathrm{d}\)?
Answer Format:
Enter your answer as an integer.
1. TRANSLATE the problem information
- Given information:
- Diameter endpoints: \(\mathrm{A(-5, 7)}\) and \(\mathrm{B(7, -11)}\)
- Need radius in form \(\mathrm{3\sqrt{d}}\)
- What this tells us: The distance between A and B gives us the diameter length
2. INFER the approach
- To find the radius, we first need the diameter length
- Distance formula will give us the diameter
- Then radius = diameter ÷ 2
3. Apply the distance formula
Using \(\mathrm{d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}\):
- \(\mathrm{\Delta x = 7 - (-5) = 12}\)
- \(\mathrm{\Delta y = -11 - 7 = -18}\)
- \(\mathrm{Diameter = \sqrt{12^2 + (-18)^2}}\)
- \(\mathrm{= \sqrt{144 + 324}}\)
- \(\mathrm{= \sqrt{468}}\)
4. SIMPLIFY the radical
- Factor 468 to find perfect square factors
- \(\mathrm{468 = 4 \times 117 = 4 \times 9 \times 13 = 36 \times 13}\)
- \(\mathrm{So \sqrt{468} = \sqrt{36 \times 13} = 6\sqrt{13}}\)
5. INFER the final step
- Since \(\mathrm{diameter = 6\sqrt{13}}\), then \(\mathrm{radius = (6\sqrt{13}) \div 2 = 3\sqrt{13}}\)
- Comparing with form \(\mathrm{3\sqrt{d}}\): \(\mathrm{d = 13}\)
Answer: 13
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students find the correct distance \(\mathrm{\sqrt{468} = 6\sqrt{13}}\) but forget that this represents the diameter, not the radius.
They incorrectly conclude that \(\mathrm{6\sqrt{13}}\) is already the radius and try to force it into the \(\mathrm{3\sqrt{d}}\) form, getting confused about what d should be. This leads to confusion and guessing rather than systematic solution.
Second Most Common Error:
Poor SIMPLIFY execution: Students calculate \(\mathrm{\sqrt{468}}\) correctly but cannot factor out the perfect squares effectively.
They might leave the answer as \(\mathrm{\sqrt{468}}\) or incorrectly simplify it (perhaps as \(\mathrm{2\sqrt{117}}\)), making it impossible to match the required \(\mathrm{3\sqrt{d}}\) format. This causes them to get stuck and abandon the systematic approach.
The Bottom Line:
This problem tests whether students can connect coordinate geometry (distance formula) with circle properties (radius-diameter relationship) while managing multi-step algebraic simplification. The key insight is recognizing that finding the distance is only the first step—you must then convert diameter to radius.