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Write the equation of the cirlcle with diameter endpoints of (-6,3) and (-14,13)

User Ddutra
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1 Answer

9 votes

Answer:

(x+10)² + (y-8)² = 41

Explanation:

First, we find the length of the diameter.

from distance formula, d = SQR ROOT OF [(-14+6)² + (13-3)²]

diameter = SQR ROOT OF [8² + 10²] = √164 = √4√41 = 2√41

***** so the radius is half that, r =√41

The center of the circle is at the midpoint of the diamater.

x = (1/2)(-6 + -14) = (1/2)(-20) = -10

y = (1/2)(3 + 13) = (1/2)(16) = 8

center (h,k) = (-10,8)

circle equation: (x-h)² + (y-k)² = r²

(x- -10)² + (y - 8)² = (√41)²

(x+10)² + (y-8)² = 41

Sorry it took a while.

User Jeo
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