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
Welcome back ~👻Masana — missed you this week. Step 1: Solve Question 1.1 M(a; b) is a point on the line y = x + 1. To write b in terms of a, substitute the coordinates of M into the equation of the line. b = a + 1 The expression for b in terms of a is b = a + 1. Step 2: Solve Question 1.2 To calculate the coordinates of M(a; b), we use the fact that M is the center of the circle and K(5; 7) and R(6; 0) are points on the circle. This means the distance from M to K is equal to the distance from M to R (both are the radius, r). MK^2 = MR^2 (a-5)^2 + (b-7)^2 = (a-6)^2 + (b-0)^2 Expand both sides: (a^2 - 10a + 25) + (b^2 - 14b + 49) = (a^2 - 12a + 36) + b^2 Subtract a^2 and b^2 from both sides: -10a - 14b + 74 = -12a + 36 Rearrange the terms: 2a - 14b + 38 = 0 Divide by 2: a - 7b + 19 = 0 (*) Substitute b = a + 1 (from Step 1) into equation (*): a - 7(a + 1) + 19 = 0 a - 7a - 7 + 19 = 0 -6a + 12 = 0 -6a = -12 a = 2 Now substitute a = 2 back into b = a + 1: b = 2 + 1 b = 3 The coordinates of M are (2; 3). Step 3: Solve Question 2 Given that the coordinates of M are (2; 3), we need to calculate the length of the radius of the circle. K(5; 7) is a point on the circle. The radius r is the distance MK. Using the distance formula: r = sqrt((x_K - x_M)^2 + (y_K - y_M)^2) r = sqrt((5 - 2)^2 + (7 - 3)^2) r = sqrt((3)^2 + (4)^2) r = sqrt(9 + 16) r = sqrt(25) r = 5 The radius of the circle is 5 units. Step 4: Solve Question 3 Determine the equation of the tangent to the circle at K(5; 7), given M(2; 3) is the center. First, find the gradient of the radius MK: m_MK = (y_K - y_M)/(x_K - x_M) m_MK = (7 - 3)/(5 - 2) m_MK = (4)/(3) The tangent line at K is perpendicular to the radius MK. The gradient of the tangent line, m_tangent, is the negative reciprocal of m_MK. m_tangent = -(1)/(m_MK) m_tangent = -(1)/(4)3 m_tangent = -(3)/(4) Now, use the point-slope form of a linear equation, y - y_1 = m(x - x_1), with point K(5; 7) and m_tangent = -(3)/(4). y - 7 = -(3)/(4)(x - 5) y - 7 = -(3)/(4)x + (15)/(4) Add 7 to both sides: y = -(3)/(4)x + (15)/(4) + 7 Convert 7 to a fraction with denominator 4: 7 = (28)/(4). y = -(3)/(4)x + (15)/(4) + (28)/(4) y = -(3)/(4)x + (43)/(4) The equation of the tangent is y = -(3)/(4)x + (43)/(4). Send me the next one 📸