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 :

Let A(1, 4) and B(1, $$-$$5) be two points. Let P be a point on the circle
(x $$-$$ 1)² + (y $$-$$ 1)² = 1 such that (PA)² + (PB)² have maximum value, then the points, P, A and B lie on :

If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x² + y² = 1 is a circle of radius r, then r is equal to :

In the circle given below, let OA = 1 unit, OB = 13 unit and PQ $$ \bot $$ OB. Then, the area of the triangle PQB (in square units) is :