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 the problems: 4.1 Give a reason why TS NP. Step 1: Identify the given information. The problem states that T(0; 5) is the center of the circle. The line through P and S(-3; 8) intersects the circle at N. It is also given that NS = PS. Step 2: Apply the relevant geometric theorem. NP is a chord of the circle. S is the midpoint of the chord NP because NS = PS. The line segment TS connects the center of the circle (T) to the midpoint of the chord (S). A line from the center of a circle to the midpoint of a chord is perpendicular to the chord. Reason: The line segment from the center of a circle to the midpoint of a chord is perpendicular to the chord. Therefore, TS NP. 4.2 Determine the equation of the line passing through N and P in the form y=mx+c. Step 1: Calculate the gradient of TS. Given points T(0; 5) and S(-3; 8). m_TS = (y_S - y_T)/(x_S - x_T) = (8 - 5)/(-3 - 0) = (3)/(-3) = -1 Step 2: Determine the gradient of NP. From question 4.1, we know that TS NP. For perpendicular lines, the product of their gradients is -1. m_NP · m_TS = -1 m_NP · (-1) = -1 m_NP = 1 Step 3: Find the equation of the line NP. The line NP passes through S(-3; 8) and has a gradient m_NP = 1. Using the point-slope form y - y_1 = m(x - x_1): y - 8 = 1(x - (-3)) y - 8 = x + 3 y = x + 3 + 8 y = x + 11 The equation of the line passing through N and P is y = x + 11. 4.3 Determine the equations of the tangents to the circle that are parallel to the x-axis. Step 1: Find the coordinates of point P. Point P is where the circle cuts the y-axis, and it lies on the line y = x + 11. For a point on the y-axis, x=0. Substitute x=0 into the equation of line NP: y = 0 + 11 y = 11 So, P(0; 11). Step 2: Calculate the radius of the circle. The center of the circle is T(0; 5) and P(0; 11) is a point on the circle. The radius r is the distance TP. r = sqrt((x_P - x_T)^2 + (y_P - y_T)^2) r = sqrt((0 - 0)^2 + (11 - 5)^2) r = sqrt(0^2 + 6^2) r = sqrt(36) r = 6 Step 3: Determine the equations of the horizontal tangents. Tangents parallel to the x-axis are horizontal lines of the form y = k. These tangents occur at the highest and lowest points of the circle. The y-coordinate of the center is y_T = 5. The radius is r = 6. The highest point is y_T + r = 5 + 6 = 11. The lowest point is y_T - r = 5 - 6 = -1. The equations of the tangents are y = 11 and y = -1. 4.4 Determine the length of MT. Step 1: Find the coordinates of point M. Point M is where the line NP intersects the x-axis. For a point on the x-axis, y=0. Substitute y=0 into the equation of line NP (y = x + 11): 0 = x + 11 x = -11 So, M(-11; 0). Step 2: Calculate the distance MT. Given points M(-11; 0) and T(0; 5). MT = sqrt((x_T - x_M)^2 + (y_T - y_M)^2) MT = sqrt((0 - (-11))^2 + (5 - 0)^2) MT = sqrt((11)^2 + (5)^2) MT = sqrt(121 + 25) MT = sqrt(146) The length of MT is sqrt(146). 4.5 Another circle is drawn through the points S, T and M. Determine, with reasons, the equation of this circle STM in the form (x-a)^2 + (y-b)^2 = r^2. Step 1: Identify the points and a key geometric property. The points are S(-3; 8), T(0; 5), and M(-11; 0). From question 4.1, we established that TS NP. Since M lies on the line NP, it means TS SM. Therefore, TSM = 90^. Step 2: Determine the diameter of the circle STM. If an angle inscribed in a circle is 90^, then the chord subtending that angle is the diameter of the circle. Since TSM = 90^, the segment TM is the diameter of the circle passing through S, T, M. Step 3: Find the center of the circle STM. The center of the circle is the midpoint of its diameter TM. Let the center be (a, b). a = (x_T + x_M)/(2) = (0 + (-11))/(2) = -(11)/(2) b = (y_T + y_M)/(2) = (5 + 0)/(2) = (5)/(2) The center of the circle is (-(11)/(2); (5)/(2)). Step 4: Calculate the radius squared (r^2) of the circle STM. The diameter is MT. From question 4.4, MT = sqrt(146). The radius r is half the diameter: r = sqrt(146)2. r^2 = (sqrt(146)2)^2 = (146)/(4) = (73)/(2) Step 5: Write the equation of the circle STM. The equation of a circle is (x-a)^2 + (y-b)^2 = r^2. Substituting the center (-(11)/(2); (5)/(2)) and r^2 = (73)/(2): (x - (-(11)/(2)))^2 + (y - (5)/(2))^2 = (73)/(2) (x + (11)/(2))^2 + (y - (5)/(2))^2 = (73)/(2) The equation of the circle STM is (x + (11)/(2))^2 + (y - (5)/(2))^2 = (73)/(2). That's 2 down. 3 left today — send the next one.