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Find an equation of the circle that stisfies the given conditions center (-1, 5); passes through (-4, -6)

0 votes
Find an equation of the circle that stisfies the given conditions
center (-1, 5); passes through (-4, -6).
asked Mar 14, 2014 in ALGEBRA 1 by angel12 Scholar

1 Answer

+1 vote

 

Circle center (h, k) = (- 1, 5) and the point is (-4, -6).

The distance between center and the point is to be the radious of circle.

Radius (r) = sqrt [(x -h)2 + (y -k)2].

Substitute the values of  (h, k) = (- 1, 5) and (x, y) = (-4, -6) in (r) = sqrt [(x -h)2 + (y -k)2].

r = sqrt [(- 4 + 1)2 + (- 6 - 5)2]

= sqrt [(- 3)2 + (- 11)2]

= sqrt [9 +121]

= sqrt [130].

Circle equation is (x -h)2 + (y -k)2 = r 2.

(x + 1)2 + (y - 5)2 = (sqrt [130]) 2

(x + 1)2 + (y - 5)2 = 130

x2 + 2x + 1 + y2 - 10y + 25 = 130

x2 + y2 + 2x - 10y - 104 = 0.

Circle equation is x2 + y2 + 2x - 10y - 104 = 0.

 

answered Mar 15, 2014 by dozey Mentor

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