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: Write the equation of the circle in the form (x-a)^2 + (y-b)^2 = r^2. The given equation of the circle is x^2 + y^2 - 4x - 2y - 13 = 0. To convert this to the standard form, we complete the square for the x and y terms. x^2 - 4x + y^2 - 2y = 13 Group the x terms and y terms: (x^2 - 4x) + (y^2 - 2y) = 13 Complete the square for x^2 - 4x by adding ((-4)/(2))^2 = (-2)^2 = 4. Complete the square for y^2 - 2y by adding ((-2)/(2))^2 = (-1)^2 = 1. Add these values to both sides of the equation: (x^2 - 4x + 4) + (y^2 - 2y + 1) = 13 + 4 + 1 (x - 2)^2 + (y - 1)^2 = 18 The equation of the circle in the required form is: (x - 2)^2 + (y - 1)^2 = 18 Step 2: Write down the coordinates of N and the length of NT. From the standard form of the circle's equation (x-a)^2 + (y-b)^2 = r^2, the center is (a,b) and the radius is r. From (x - 2)^2 + (y - 1)^2 = 18: The coordinates of the centre N are (2,1). The radius squared is r^2 = 18. The length of the radius NT is r = sqrt(18). Simplify sqrt(18): sqrt(18) = sqrt(9 × 2) = 3sqrt(2). The coordinates of N are (2,1) and the length of NT is 3sqrt(2). Step 3: Determine the equation of the tangent QT in the form y = mx + c. We are given that TNP is a diameter and P is (5,4). N is the center (2,1). Since N is the midpoint of the diameter TP, we can find the coordinates of T using the midpoint formula: Let T = (x_T, y_T). x_N = (x_T + x_P)/(2) 2 = (x_T + 5)/(2) 4 = x_T + 5 x_T = -1 y_N = (y_T + y_P)/(2) 1 = (y_T + 4)/(2) 2 = y_T + 4 y_T = -2 So, the coordinates of T are (-1, -2). The tangent QT is perpendicular to the radius NT at the point of tangency T. First, calculate the gradient of the radius NT (m_NT): N = (2,1) and T = (-1,-2). m_NT = (y_N - y_T)/(x_N - x_T) = (1 - (-2))/(2 - (-1)) = (1+2)/(2+1) = (3)/(3) = 1 Since QT is perpendicular to NT, the gradient of the tangent QT (m_QT) is the negative reciprocal of m_NT: m_QT = -(1)/(m_NT) = -(1)/(1) = -1 Now, use the point-slope form of a linear equation y - y_1 = m(x - x_1) with point T (-1,-2) and m_QT = -1: y - (-2) = -1(x - (-1)) y + 2 = -1(x + 1) y + 2 = -x - 1 y = -x - 1 - 2 y = -x - 3 The equation of the tangent QT is y = -x - 3. Step 4: Determine the coordinates of S. The circle with centre S(a; b) touches the circle with centre N externally at T. QT is a tangent to both circles. This means that the radius ST is also perpendicular to QT at T. Since NT is perpendicular to QT and ST is perpendicular to QT at the same point T, the points N, T, and S must be collinear. Also, since the circles touch externally at T, T lies between N and S. We are given NS = 3NT. We know NT is the radius of the first circle, NT = 3sqrt(2). So, NS = 3 × (3sqrt(2)) = 9sqrt(2). Since N, T, S are collinear and T is between N and S, the distance NS = NT + TS. Therefore, 3NT = NT + TS TS = 2NT. This means that T divides the line segment NS in the ratio NT:TS = 1:2. We can use the section formula or simply observe that NS = 3 NT if N was the origin, but here T is between N and S. A more direct way is to use the fact that T is a point on the line segment NS such that NT = (1)/(3) NS. This means that NT = (1)/(3) NS. So, S - N = 3(T - N). S = 3T - 3N + N S = 3T - 2N. Using the coordinates: S_x = 3x_T - 2x_N = 3(-1) - 2(2) = -3 - 4 = -7. S_y = 3y_T - 2y_N = 3(-2) - 2(1) = -6 - 2 = -8. The coordinates of S are (-7, -8). Got more? Send 'em!