Equation of Tangent to a Circle
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
\[y=-\frac{3}{4}x +\frac{9}{4}+2=-\frac{3}{4}x +\frac{17}{4}\]
