Given the equation x ^ 2 + y ^ 2 + KX + 2Y + k = 0 of a circle, if there are two tangents of the circle made by passing through the fixed point P (1, - 1), then the condition that K should satisfy is

Given the equation x ^ 2 + y ^ 2 + KX + 2Y + k = 0 of a circle, if there are two tangents of the circle made by passing through the fixed point P (1, - 1), then the condition that K should satisfy is

It can be seen from the question that P must be outside the circle
So 1 ^ 2 + (- 1) ^ 2 + K-2 + k > 0
The solution is k > 0