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.
ScanSolve AI Answer
Step 1: Determine the coordinates of M and the radius of the circle (for 7.1) The equation of the circle with centre M is given as (x+2)^2 + (y-1)^2 = r^2. Comparing this to the standard form of a circle equation (x-a)^2 + (y-b)^2 = r^2, the coordinates of the centre M are (-2; 1). The point S(1; -2) lies on the circle. Substitute these coordinates into the equation to find r^2: (1+2)^2 + (-2-1)^2 = r^2 (3)^2 + (-3)^2 = r^2 9 + 9 = r^2 r^2 = 18 The radius r = sqrt(18) = sqrt(9 × 2) = 3sqrt(2). The coordinates of M are (-2; 1) and the radius is 3sqrt(2). Step 2: Determine the equation of the tangent RS (for 7.2) The tangent RS touches the circle at point S(1; -2). The radius MS is perpendicular to the tangent at S. First, calculate the gradient of the radius MS, m_MS, using M(-2; 1) and S(1; -2): m_MS = (y_S - y_M)/(x_S - x_M) = (-2 - 1)/(1 - (-2)) = (-3)/(1 + 2) = (-3)/(3) = -1 Since the tangent RS is perpendicular to the radius MS, the gradient of the tangent, m_RS, is the negative reciprocal of m_MS: m_RS = -(1)/(m_MS) = -(1)/(-1) = 1 Now, use the point-slope form of a linear equation, y - y_1 = m(x - x_1), with point S(1; -2) and m_RS = 1: y - (-2) = 1(x - 1) y + 2 = x - 1 y = x - 1 - 2 The equation of the tangent RS is y = x - 3. Step 3: Determine the coordinates of point P (for 7.3) The circles with centres P and M touch externally at point S. This means S lies on the line segment MP. We are given the ratio MS : MP = 1 : 3. This implies that the distance from M to S is one-third of the distance from M to P. Let M be (-2; 1) and S be (1; -2). Let P be (a; b). We can use the section formula or vector approach. Using the vector approach: The vector MS represents the displacement from M to S: MS = (x_S - x_M, y_S - y_M) = (1 - (-2), -2 - 1) = (3, -3) Since MS : MP = 1 : 3, the vector MP is 3 times the vector MS: MP = 3 × MS = 3(3, -3) = (9, -9) The vector MP also represents the displacement from M to P: MP = (x_P - x_M, y_P - y_M) = (a - (-2), b - 1) = (a+2, b-1) Equating the components of MP: a+2 = 9 a = 7 b-1 = -9 b = -8 The coordinates of point P are (7; -8). 3 done, 2 left today. You're making progress.