ACT Math Practice Question #1284 (Medium (30)) | Test Citadel
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ACT Math Difficulty: Medium (30)

Digital ACT Math Practice Question #1284

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A circle in the standard $(x,y)$-coordinate plane has the equation $x^2 + y^2 - 6x + 8y - 11 = 0$. What is the length of the radius of this circle?
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$r = \sqrt{(d/2)^2 + (e/2)^2 - f} = \sqrt{(-3)^2 + (4)^2 - (-11)} = \sqrt{9 + 16 + 11} = \sqrt{36} = 6$.

Masterclass Solution & Distractor Trap Analysis

Complete the square for $x$ and $y$: $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16$ $(x - 3)^2 + (y + 4)^2 = 36$. The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$. Here, $r^2 = 36$, so the radius is $r = \sqrt{36} = 6$....

Distractor Analysis: Trap choice eliminates careless test-takers who confuse roots with coordinates...

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