The points (−3, 5) and (7, 9) are the ends of the diameter of a circle. a) Find the equation of the circle. b) Find the equation of the normal to the circle at (3, 4).

Here are the solutions to the questions. Question 11: The points (-3, 5) and (7, 9) are the ends of the diameter of a circle. a) Find the equation of the circle. Step 1: Find the coordinates of the center of the circle (h, k). The center is the midpoint of the diameter. Let (x_1, y_1) = (-3, 5) and (x_2, y_2) = (7, 9). h = (x_1 + x_2)/(2) = (-3 + 7)/(2) = (4)/(2) = 2 k = (y_1 + y_2)/(2) = (5 + 9)/(2) = (14)/(2) = 7 The center of the circle is (2, 7). Step 2: Find the radius squared (r^2) of the circle. The radius is the distance from the center to one of the endpoints of the diameter. We'll use the center (2, 7) and the point (-3, 5). r^2 = (x - h)^2 + (y - k)^2 r^2 = (-3 - 2)^2 + (5 - 7)^2 r^2 = (-5)^2 + (-2)^2 r^2 = 25 + 4 r^2 = 29 The radius squared is 29. Step 3: Write the equation of the circle using the standard form (x - h)^2 + (y - k)^2 = r^2. (x - 2)^2 + (y - 7)^2 = 29 b) Find the equation of the normal to the circle at (3, 4). Step 1: Understand that the normal to a circle at any point on its circumference is the line that passes through that point and the center of the circle. The point on the circle is (3, 4). The center of the circle is (2, 7) (from part a). Step 2: Calculate the gradient of the normal (m_n) using the two points (3, 4) and (2, 7). m_n = (y_2 - y_1)/(x_2 - x_1) = (7 - 4)/(2 - 3) m_n = (3)/(-1) m_n = -3 Step 3: Use the point-slope form of a linear equation, y - y_1 = m(x - x_1), with the point (3, 4) and the gradient m_n = -3. y - 4 = -3(x - 3) Step 4: Simplify and rearrange the equation into the general form Ax + By + C = 0. y - 4 = -3x + 9 3x + y - 4 - 9 = 0 3x + y - 13 = 0 What's next?