This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to question 2: a) The equation of the circle C_3 which passes through the center of C_1 and through the points of intersection of C_1 and C_2. Step 1: Find the center of circle C_1. The general equation of a circle is x^2 + y^2 + 2gx + 2fy + c = 0, with center (-g, -f). For C_1: x^2 + y^2 - 6x - 4y + 9 = 0: 2g = -6 g = -3 2f = -4 f = -2 The center of C_1 is (3, 2). Step 2: Find the equation of the common chord (line of intersection) of C_1 and C_2. The equation of the common chord is C_1 - C_2 = 0. (x^2 + y^2 - 6x - 4y + 9) - (x^2 + y^2 - 2x - 6y + 9) = 0 -6x - 4y + 9 + 2x + 6y - 9 = 0 -4x + 2y = 0 2y = 4x y = 2x This is the equation of the line L passing through the intersection points. Step 3: Form the general equation of circle C_3. A circle passing through the intersection of C_1=0 and the line L=0 can be written as C_1 + L = 0. C_3: (x^2 + y^2 - 6x - 4y + 9) + (y - 2x) = 0 Step 4: Use the condition that C_3 passes through the center of C_1. Substitute the coordinates of the center of C_1, which is (3, 2), into the equation for C_3: (3^2 + 2^2 - 6(3) - 4(2) + 9) + (2 - 2(3)) = 0 (9 + 4 - 18 - 8 + 9) + (2 - 6) = 0 (22 - 26) + (-4) = 0 -4 - 4 = 0 -4 = 4 = -1 Step 5: Substitute = -1 back into the equation for C_3. (x^2 + y^2 - 6x - 4y + 9) - 1(y - 2x) = 0 x^2 + y^2 - 6x - 4y + 9 - y + 2x = 0 x^2 + y^2 - 4x - 5y + 9 = 0 The equation of circle C_3 is x^2 + y^2 - 4x - 5y + 9 = 0. b) The equation of the two tangents from the origin to C_1 and the length of each tangent. Step 1: Find the center and radius of C_1. For C_1: x^2 + y^2 - 6x - 4y + 9 = 0: Center (h, k) = (3, 2). Radius r = sqrt(g^2 + f^2 - c) = sqrt((-3)^2 + (-2)^2 - 9) = sqrt(9 + 4 - 9) = sqrt(4) = 2. Step 2: Calculate the length of the tangent from the origin (0,0) to C_1. The length of the tangent from an external point (x_1, y_1) to a circle x^2 + y^2 + 2gx + 2fy + c = 0 is sqrt(x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c). For the origin (0,0) and C_1: L = sqrt(0^2 + 0^2 - 6(0) - 4(0) + 9) L = sqrt(9) L = 3 The length of each tangent is 3 units. Step 3: Find the equations of the tangents. Let the equation of a tangent line passing through the origin $(0,