How do you work out the tangent to a circle at a given point on the circle?

(Suppose the coordinates of the point on the circle is (2, 3))To calculate the tangents equation we will first work out it's gradient. The tangent to a circle at a point is perpendicular to the radius of the circle at that point (here i would have drawn and labelled the circle, the point on the circle, the circles centre and highlighted the circle's radius to this point. I would also draw the tangent to this point and highlight the fact it is indeed perpendicular). We can work out the gradient of the tangent by first working out the gradient of this line (the one highlighted). The gradient of a line between two points is the change in y/change in x. The centre of the circle is at (0, 0). The gradient of the radius is therefor (3-0)/(2-0) = 3/2. The gradient of a line perpendicular to a line with a known equation is -1/known gradient. The gradient of the tangent is therefor -1/(3/2) = -2/3. The equation of this line will be of the form y = mx + c. We know m (the gradient) and now need to work out c (the y intercept). We know the line intercepts the point (2, 3), so we can put the x and y coordinates into our equation and solve for c. 3 = (-2/3) x (2) + c, 3 = -4/3 + c, 3 + 4/3 = c, c = 13/3. The equation of our tangent is therefor y = -2/3x + 13/3.