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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You're on a roll — Here are the solutions to the circle geometry questions: 7. The diagram below shows a circle with center P(5,5) and radius 5 units. a) Write down in terms of x and y the equation of the circle in the form ax^2 + by^2 + cx + dy + e = 0 where a, b, c, d and e are constants. (2mks) Step 1: Use the standard equation of a circle. The standard equation of a circle with center (h, k) and radius r is (x-h)^2 + (y-k)^2 = r^2. Given center P(5,5), so h=5 and k=5. Given radius r=5. Step 2: Substitute the values into the standard equation. (x-5)^2 + (y-5)^2 = 5^2 Step 3: Expand and rearrange the equation into the form ax^2 + by^2 + cx + dy + e = 0. (x^2 - 10x + 25) + (y^2 - 10y + 25) = 25 x^2 + y^2 - 10x - 10y + 25 + 25 - 25 = 0 x^2 + y^2 - 10x - 10y + 25 = 0 Here, a=1, b=1, c=-10, d=-10, e=25. b) Determine the gradient of PQ (1m) c) Find the equation of the tangent at Q in the form ax + by = c (3mks) Note: The diagram mentioned in the question is not provided. Without the diagram, the specific coordinates of point Q cannot be determined. Therefore, parts b) and c) cannot be solved to a numerical answer. Below is the general method for solving these parts if the coordinates of Q were known. General Method for b) and c): Step 1: Identify the coordinates of Q. Let's assume, for demonstration purposes, that point Q has coordinates (x_Q, y_Q). (These coordinates would be given in the missing diagram). The center of the circle is P(5,5). Step 2: Calculate the gradient of the radius PQ (for part b). The gradient m_PQ is given by the formula: m_PQ = (y_Q - y_P)/(x_Q - x_P) = (y_Q - 5)/(x_Q - 5) This would yield a specific numerical value if (x_Q, y_Q) were known. Step 3: Calculate the gradient of the tangent at Q (for part c). The tangent at any point on a circle is perpendicular to the radius at that point. If m_PQ is the gradient of the radius PQ, then the gradient of the tangent, m_tangent, is the negative reciprocal of m_PQ: m_tangent = -(1)/(m_PQ) Step 4: Find the equation of the tangent at Q (for part c). Using the point-slope form of a linear equation, y - y_1 = m(x - x_1), with point Q(x_Q, y_Q) and gradient m_tangent: y - y_Q = m_tangent(x - x_Q) This equation would then be rearranged into the form ax + by = c. What's next?