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
Fresh day 🩷🫶kEaMoW🩷🥹, let's solve. To find the equation of the tangent and normal to the circle, we first need to determine the center of the circle and the gradient of the radius to the given point. Step 1: Find the center of the circle. The given equation of the circle is x^2 + 4x + y^2 - 6y = 91. We complete the square to find the standard form (x-h)^2 + (y-k)^2 = r^2. (x^2 + 4x + 4) + (y^2 - 6y + 9) = 91 + 4 + 9 (x+2)^2 + (y-3)^2 = 104 The center of the circle is C(-2, 3). The given point on the circle is P(0, -7). Step 2: Find the gradient of the radius (m_radius) connecting the center C(-2, 3) and the point P(0, -7). The formula for the gradient between two points (x_1, y_1) and (x_2, y_2) is m = (y_2 - y_1)/(x_2 - x_1). m_radius = (-7 - 3)/(0 - (-2)) = (-10)/(2) = -5 Step 3: Find the gradient of the tangent (m_tangent). The tangent line is perpendicular to the radius at the point of tangency. If two lines are perpendicular, the product of their gradients is -1. m_tangent = -(1)/(m_radius) = -(1)/(-5) = (1)/(5) Step 4: Find the equation of the tangent line. Using the point-slope form y - y_1 = m(x - x_1) with point P(0, -7) and m_tangent = (1)/(5): y - (-7) = (1)/(5)(x - 0) y + 7 = (1)/(5)x y = (1)/(5)x - 7 Step 5: Find the gradient of the normal (m_normal). The normal line is the line that passes through the center of the circle and the point of tangency. Therefore, its gradient is the same as the gradient of the radius. m_normal = m_radius = -5 Step 6: Find the equation of the normal line. Using the point-slope form y - y_1 = m(x - x_1) with point P(0, -7) and m_normal = -5: y - (-7) = -5(x - 0) y + 7 = -5x y = -5x - 7 The equation of the tangent is y = (1)/(5)x - 7. The equation of the normal is y = -5x - 7. Send me the next one 📸