In die diagram hieronder lê A, B (-5; 12) en C (u; 0) op die sirkel met middelpunt O by die oorsprong - NSC Technical Mathematics - Question 2 - 2021 - Paper 1

From the equation of the circle:

t = rac{y}{x}

At point B (–5, 12), substituting the y and x values gives:

t = rac{12}{-5} = -rac{12}{5}

To find the equation of the tangent line at point B (-5, 12), we first need the gradient (m) using the derived radius:

m_{OB} = rac{y_2 - y_1}{x_2 - x_1} = rac{12 - 0}{-5 - 0} = rac{12}{-5} = -rac{12}{5}

The tangent line has a slope that is the negative reciprocal, thus:

m_{tangent} = rac{5}{12}

Using the point-slope form of the line equation:

$y - y_1 = m(x - x_1)$

Substituting B (-5, 12):

y - 12 = rac{5}{12}(x + 5)

Simplifying gives:

y = rac{5}{12}x + rac{25}{12} + 12 = rac{5}{12}x + rac{169}{12}