A circle has equation: (x - 2)^2 + (y - 2)^2 = 16. It intersects the y-axis (y > 0) at point P and the x-axis (x < 0) at point Q. Find the equation of the line connecting P and Q and of the line perpendicular to PQ passing through the circle's centre.

(x - 2)² + (y - 2)² = 16 x = 0 --> 4 + (y - 2)² = 16 --> y - 2 = ±2sqrt{2} --> y = 2 ± 2sqrt{2} --> P(0,2 + 2sqrt{2})y = 0 --> (x - 2)² + 4 = 16 --> x - 2 = ±2sqrt{2} --> x = 2 ± 2sqrt{2} --> Q(2 - 2sqrt{2},0) Line PQ: Gradient = (0 - (2 + 2sqrt{2}))/((2 - 2sqrt{2}) - 0) = 3 + 2sqrt{2} Equation of line PQ: y = (3 + 2sqrt{2})x + (2 + 2sqrt{2}) Line perpendicular to PQ passing through the circle's centre: Gradient = -1/(3 + 2sqrt{2}) = 2sqrt{2} - 3 Equation of line: y - 2 = (2sqrt{2} - 3)(x - 2) --> y = (2sqrt{2} - 3)x + (8 - 4sqrt{2})