Let C₁ and C₂ be the centres of the circles x² + y² – 2x – 2y – 2 = 0 and x² + y² – 6x – 6y + 14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC₁QC₂ is :

The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :

A square is inscribed in the circle x² + y² – 6x + 8y – 103 = 0 with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is :

Two circles with equal radii are intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :