A predator moves in a circle of radius \( \sqrt{2} \) centre \((0,0)\), while a prey moves along the line \(y = x\). If \(0 \le x \le 2\), at which point(s)...

Answer Details

The predator moves in a circular path with a radius of √2 centered at the origin (0,0), while the prey moves along the line y = x. The prey's movement is restricted between x = 0 and x = 2. At some point, the predator will catch up to the prey, and they will meet. The predator will catch the prey when the distance between them is less than or equal to √2. We can find the equation of the circle centered at (0,0) with a radius of √2: x^2 + y^2 = (√2)^2 x^2 + y^2 = 2 We can substitute y = x into the equation to get: x^2 + x^2 = 2 2x^2 = 2 x^2 = 1 x = ±1 Therefore, the possible points of intersection are (1,1) and (-1,-1), but we know that the prey can only be at x values between 0 and 2, so the only possible point of intersection is (1,1). Therefore, the answer is (1,1) only.