Question

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

Answer 1

  

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.