If P is a point on the circle x^2 + y^2 - 2x - 4y + 1 = 0 , Q is a point on the circle x^2 + y^2 = 13 and x - y + 2 = 0 is the perpendicular bisector of PQ , then the sum of the squares of the distances of all such points P from the origin is _ _ _ _ .

Step-by-step solution

Let P (h, k) be a point on the first circle. Then, h^2 + k^2 - 2h - 4k + 1 = 0 ... (1) The reflection of P across the line x - y + 2 = 0 is Q . x - h 1 = y - k -1 = -2 h - k + 2 1^2 + (-1)^2 = -h + k - 2 x = k - 2 and y = h + 2 So, Q (k - 2, h + 2) . Since Q lies on the second circle x^2 + y^2 = 13 , (k - 2)^2 + (h + 2)^2 = 13 h^2 + k^2 + 4h - 4k - 5 = 0 ... (2) Subtracting equation (1) from equation (2), we get: 6h - 6 = 0 h = 1 Substituting h = 1 into equation (1): 1^2 + k^2 - 2(1) - 4k + 1 = 0 k^2 - 4k = 0 k = 0