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If the endpoints of the diameter of a circle are (−10, −8) and (−6, −2), what is the standard form equation of the circle? A) (x − 8)2 + (y − 5)2 = 13 B) (x + 8)2 + (y + 5)2 = 13 C) (x − 8)2 + (y − 5)2 = 13 D) (x + 8)2 + (y + 5)2 = 13

User Kazunori
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2 Answers

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Answer: D) (x+8)2 +( y+5)2 = 13

Explanation:

Got this right on USA test prep

User MylesK
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Selections B and D both appear to be appropriate.
B) (x+8)² + (y+5)² = 13
D) (x+8)² + (y+5)² = 13

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The center is at the midpoint of the diameter, ((-10-6)/2, (-8-2)/2) = (-8, -5). For center (h, k) and radius r, the equation is
(x -h)² +(y -k)² = r²
(x -(-8))² +(y -(-5))² = (√13)²
(x+8)² + (y+5)² = 13
User Thamara
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