JEE Math Practice Question

Let the lengths of intercepts on x-axis and y-axis made by the circle x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2$${\sqrt 2 }$$ and 2$${\sqrt 5 }$$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :

1. A.$${\sqrt {10} }$$
2. B.$${\sqrt {6} }$$
3. C.$${\sqrt {11} }$$
4. D.$${\sqrt {7} }$$