In the diagram below, O is the centre of the circle at the origin - NSC Technical Mathematics - Question 2 - 2024 - Paper 2

To find the equation of the tangent at point M (-5, 12), we first find the slope of the radius OM: $m_{OM} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 0}{-5 - 0} = -\frac{12}{5}$ The slope of the tangent ML (perpendicular to OM) is the negative reciprocal: $m_{ML} = \frac{5}{12}$ Using point-slope form: $y - y_1 = m(x - x_1)$ Substituting M (-5, 12): $y - 12 = \frac{5}{12}(x + 5)$ Expanding this gives:

The y-intercept L occurs when x = 0 in the equation of tangent ML: Substituting x = 0 into the tangent equation: $y = \frac{5}{12}(0) + \frac{169}{12} = \frac{169}{12}$ Thus, the coordinates of point L are: $L(0, \frac{169}{12})$

The angle β is the angle of inclination of line MD. First, find the slope of line MD: $m_{MD} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 0}{-5 - (-3)} = \frac{12}{-2} = -6$ Using the tangent function to find β: $\tan(β) = |m_{MD}| = 6$ Thus, the angle β is: $β = \arctan(6) \approx 80.54^ ext{°}$