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As the tangent is a straight line, the equation of the tangent will be of the form \(y = mx + c\). We can use perpendicular gradients to find the value of \(m\), then use the coordinates of P to find the value of \(c\) in the equation.

Find the equation of the tangent to the circle \(x^2 + y^2 = 25 \) at the point (3, -4).

The tangent will have an equation in the form \(y = mx + c\) so to find the equation you need to find the values of \(m\) and \(c\). First, find \(m\), the gradient of the tangent.

On a diagram, draw the circle and the tangent at the point P (3, -4) and draw the radius from the centre (0, 0) to the point P. The gradient of the radius is given by \(\frac{change~in~y}{change~in~x} =\frac{-4}{3}\)

The radius that joins the centre of the circle (0, 0) to the point P is at right angles to the tangent, so the gradient of the tangent is the negative reciprocal of the gradient of the radius. This means that the gradient of the tangent, \(m = \frac{3}{4}\).

Next, find the value of \(c\). For the point P, the value of \(x = 3\) and the value of \(y = -4\). Substituting these values, as well as the value of \(m = \frac{3}{4}\), in to the formula \(y = mx + c\) gives:

\(y = mx + c\)

\(-4 = \frac{3}{4} \times 3 + c\)

\(-4 = \frac{9}{4} + c\)

\(-4 - \frac{9}{4} = c\)

\(c = -\frac{25}{4}\)

So the equation of the tangent to the circle \(x^2 + y^2 = 25 \) at the point (3, -4) is \(y = \frac{3}{4}x - \frac{25}{4}\)