The diagram shows a circle with centre (0, 0) and a tangent at the point (−12, 16) - OCR - GCSE Maths - Question 14 - 2019 - Paper 4

The tangent line is perpendicular to the radius. The gradient of the tangent is the negative reciprocal of the gradient of the radius:

$m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4}$

Using the point-slope form of the line equation:

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

Substituting the point \((−12, 16)\) and the gradient of the tangent:

$y - 16 = \frac{3}{4}(x + 12)$

Expanding this gives:

$y - 16 = \frac{3}{4}x + 9$

Therefore:

$y = \frac{3}{4}x + 25$

To find the point where the tangent crosses the y-axis, set (x = 0):

$y = \frac{3}{4}(0) + 25 = 25$

Thus, we find that (p = 25).