That will never happen for the group used in Bitcoin. That specific point, is not a point on the elliptic curve used by Bitcoin. Such a point, call it Q, would have order 2, i.e. 2Q = O, the point at infinity. However, the points on the elliptic curve Y^2 = X^3 + 7 over Fp form a cyclic group of order a huge prime (almost as large as p). Hence not divisible by 2 and therefore no elements of order 2.
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