SAT Math: A Circle Has Diameter Endpoints at (-3, 2) and (5, 8), Find Its Equation
Center is midpoint (1, 5). Radius squared is distance squared = (5 - 1)^2 + (8 - 5)^2 = 16 + 9 = 25. Equation: (x - 1)^2 + (y - 5)^2 = 25.
Exact Question Text
A circle in the xy-plane has a diameter with endpoints at (-3, 2) and (5, 8). What is the standard equation of the circle?
1. Step-by-Step Formal Solution
- Center = ((-3+5)/2, (2+8)/2) = (1, 5).
- Radius r = sqrt((5 - 1)^2 + (8 - 5)^2) = sqrt(25) = 5.
- Equation: (x - 1)^2 + (y - 5)^2 = 25.
2. Fast Desmos / Calculator Shortcut
Plot points (-3, 2) and (5, 8) with midpoint (1, 5). Type (x - 1)^2 + (y - 5)^2 = 25 to verify it passes through both endpoints.
3. Distractor Trap Analysis
- Trap A (Sign Flip): Forgetting to reverse inequality or dropping negative sign.
- Trap B (Partial Evaluation): Solving for intermediate variable and stopping early.
- Trap C (Extreme Assumption): Over-inferring beyond textual evidence.
- Option Correct: Accurately satisfies all algebraic and logical constraints.
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Center is midpoint (1, 5). Radius squared is distance squared = (5 - 1)^2 + (8 - 5)^2 = 16 + 9 = 25. Equation: (x - 1)^2 + (y - 5)^2 = 25.
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