Suppose we have a point

\[(x_1,y_1)\]

on a circle and a tangent at that point. The centre of the circle is at the point \[(x_{CENTRE},y_{CENTRE})\]

say, so the gradient of the this particular radius is \[m_{RADIUS}=\frac{y_1-y_{CENTRE}}{x_1-x_{CENTRE}}\]

and the tangent, being at right angles to the radius at the point where they meet on the circle, has gradient \[m_{TANGENT}=- \frac{1}{m_{RADIUS}}\]

.The equation of the tangent is then

\[y-y_1 =m_{TANGENT}(x-x_1)\]

Example: A circle has centre

\[(3,2)\]

. Find the equation of the tangent to the circle at the point \[(6,6)\]

.The gradient of the radius drawn from

\[(3,2)\]

to \[(6,6)\]

is \[m_{RADIUS}=\frac{6-2}{6-3}= \frac{4}{3}\]

.The gradient of the tangent is then

\[m_{TANGENT} = -\frac{1}{4/3}=- \frac{3}{4}\]

.The equation of the tangent is

\[y-2=- \frac{3}{4} (x-3)=- \frac{3}{4} x+\frac{9}{4}\]

.Hence

\[y=-\frac{3}{4}x +\frac{9}{4}+2=-\frac{3}{4}x +\frac{17}{4}\]