Question

If the equation of a circle is (x - 2)2 + (y - 6)2 = 4, does it pass through point (5, 6)?

Answer 1

(x - 2)^2 + (y - 6)^2 = 4



(5 - 2)^2 + (6 - 6)^2 = 4 .... Plug in x=5 and y=6



(3)^2 + (0)^2 = 4



9 - 0 = 4



9 = 4 ... this equation is false, so (5,6) is NOT a solution



Therefore, (5,6) is NOT on the circle