Determine the centre and radius of the second circle.

Step 1: Determine the centre and radius of the second circle. The equation of the second circle is given by x^2 + y^2 + x - 3y + 2 = 0. Comparing this to the general form of a circle x^2 + y^2 + 2gx + 2fy + c = 0, we have: 2g = 1 g = (1)/(2) 2f = -3 f = -(3)/(2) c = 2 The coordinates of the centre are (-g, -f). Centre = (-(1)/(2), -(-(3)/(2))) = (-(1)/(2), (3)/(2)). The radius R is given by sqrt(g^2 + f^2 - c). R = sqrt(((1)/(2))^2 + (-(3)/(2))^2 - 2) R = sqrt((1)/(4) + (9)/(4) - 2) R = sqrt((10)/(4) - (8)/(4)) R = sqrt((2)/(4)) R = sqrt((1)/(2)) R = (1)/(sqrt(2)) = sqrt(2)2 The centre of the circle is (-(1)/(2), (3)/(2)) and the radius is sqrt(2)2. Step 2: Calculate the coordinates of A. The common chord AB has the equation y = -x + 1. Point A is an intersection of this line and the second circle x^2 + y^2 + x - 3y + 2 = 0. Substitute y = -x + 1 into the circle equation: x^2 + (-x+1)^2 + x - 3(-x+1) + 2 = 0 x^2 + (x^2 - 2x + 1) + x + 3x - 3 + 2 = 0 2x^2 + 2x = 0 2x(x+1) = 0 This gives two possible x-coordinates: x = 0 or x = -1. If x = 0, then y = -0 + 1 = 1. This gives the point (0, 1). If x = -1, then y = -(-1) + 1 = 1 + 1 = 2. This gives the point (-1, 2). From the problem description, the first circle touches the y-axis at point B. The point (0,1) lies on the y-axis. Therefore, B is (0,1). Since A and B are the intersection points, A must be the other point. The coordinates of A are (-1, 2). Step 3: Show the equation of the circle centred at M. The first circle is centred at M and touches the x-axis at C and the y-axis at B. From Step 2, we found B is (0,1). Since the circle touches the y-axis at (0,1), its radius must be r=1. Also, the y-coordinate of the centre M must be 1. Since the circle touches the x-axis at C, and from the diagram C is on the negative x-axis, the x-coordinate of the centre M must be -r = -1. So, the centre M is (-1, 1) and the radius is r=1. The equation of a circle with centre (h, k) and radius r is (x-h)^2 + (y-k)^2 = r^2. Substituting M (-1, 1) and r=1: (x - (-1))^2 + (y - 1)^2 = 1^2 (x+1)^2 + (y-1)^2 = 1 x^2 + 2x + 1 + y^2 - 2y + 1 = 1 x^2 + y^2 + 2x - 2y + 1 = 0 This matches the required equation. Step 4: Show the given equation for the tangent line. The straight line y = -x + k is a tangent to the circle with centre M, which is $x^2 + y^