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
Alright 💸🚸Kayy — let's do this. Step 1: Determine the equation of the circle (3.1). The center of the circle is M(-2; 3), so a = -2 and b = 3. The circle passes through point A(1; -1). The radius squared, r^2, is the distance squared between M and A: r^2 = (x_A - x_M)^2 + (y_A - y_M)^2 r^2 = (1 - (-2))^2 + (-1 - 3)^2 r^2 = (3)^2 + (-4)^2 r^2 = 9 + 16 r^2 = 25 The equation of the circle is: (x - (-2))^2 + (y - 3)^2 = 25 (x + 2)^2 + (y - 3)^2 = 25 Step 2: Write down the coordinates of C (3.2). The center of the circle is M(-2; 3) and the radius is r = sqrt(25) = 5. The line BC is parallel to the y-axis, meaning it is a vertical line. Since BC is tangent to the circle at C, the radius MC is perpendicular to BC. As BC is vertical, MC must be horizontal. This means the y-coordinate of C is the same as the y-coordinate of M, which is 3. From the diagram, C is to the left of M. The x-coordinate of C is x_M - r. x_C = -2 - 5 = -7 The coordinates of C are: C(-7; 3) Step 3: Determine the equation of the tangent AB (3.3). The tangent AB touches the circle at A(1; -1). The center is M(-2; 3). First, find the gradient of the radius MA: m_MA = (y_A - y_M)/(x_A - x_M) = (-1 - 3)/(1 - (-2)) = (-4)/(3) Since the tangent AB is perpendicular to the radius MA, the product of their gradients is -1: m_AB · m_MA = -1 m_AB · ((-4)/(3)) = -1 m_AB = (3)/(4) Now use the point-slope form y - y_1 = m(x - x_1) with point A(1; -1) and m_AB = (3)/(4): y - (-1) = (3)/(4)(x - 1) y + 1 = (3)/(4)x - (3)/(4) y = (3)/(4)x - (3)/(4) - 1 y = (3)/(4)x - (7)/(4) The equation of the tangent AB is: y = (3)/(4)x - (7)/(4) Step 4: Determine the length of BC (3.4). From Step 2, C(-7; 3). The line BC is a vertical line, so the x-coordinate of B is the same as C, which is -7. Point B is the intersection of the tangent AB (from Step 3) and the line x = -7. Substitute x = -7 into the equation of AB: y_B = (3)/(4)(-7) - (7)/(4) y_B = (-21)/(4) - (7)/(4) y_B = (-28)/(4) y_B = -7 So, B(-7; -7). The length of BC is the absolute difference in their y-coordinates: BC = |y_B - y_C| = |-7 - 3| = |-10| = 10 The length of BC is: 10 units Step 5: Determine the equation of the circle centred at A that has the x- and y-axis as tangents (3.5). The center of this new circle is A(1; -1). For a circle to be tangent to both the x-axis and the y-axis, its radius must be equal to the absolute value of its x-coordinate and y-coordinate. Radius r = |x_A| = |1| = 1. Radius r = |y_A| = |-1| = 1. So, the radius of this circle is r = 1. The equation of the circle is (x - x_A)^2 + (y - y_A)^2 = r^2: (x - 1)^2 + (y - (-1))^2 = 1^2 (x - 1)^2 + (y + 1)^2 = 1 The equation of the circle is: (x - 1)^2 + (y + 1)^2 = 1 Step 6: Determine all possible values of p (3.6). The original circle (Circle M) has center M(-2; 3) and radius r_M = 5 (from Step 1). The new circle (Circle N) has center $N(p