The centre of a circle is the point with coordinates (-1, 3)\nThe point A with coordinates (6, 8) lies on the circle.\nFind an equation of the tangent to the circle at A.\nGive your answer in the form $ax + by + c = 0$ where a, b and c are integers. - Edexcel - GCSE Maths - Question 21 - 2022 - Paper 1

The tangent at point A is perpendicular to the radius. The gradient of a line that is perpendicular to another is the negative reciprocal of the gradient of the other line. Thus:

$m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{\frac{5}{7}} = -\frac{7}{5}$

To find the equation of the tangent line at the point A(6, 8), we can use the point-slope form:

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

Substituting in the values, we get:

$y - 8 = -\frac{7}{5}(x - 6)$

This simplifies to:

$y - 8 = -\frac{7}{5}x + \frac{42}{5}$\n\ny = -\frac{7}{5}x + \frac{42}{5} + 8$\n\ny = -\frac{7}{5}x + \frac{42}{5} + \frac{40}{5}$\n\ny = -\frac{7}{5}x + \frac{82}{5}$$

Next, to express the equation in the form $ax + by + c = 0$, we first rearrange the equation:

Multiply through by 5 to eliminate the fraction:

$5y = -7x + 82$ \nReorganizing leads to:

$7x + 5y - 82 = 0$ \nThus, the final tangent equation is:

$7x + 5y - 82 = 0$