Suppose I have the following circle
$$
(x – x_0)^2 + (y – y_0)^2 = r^2
$$
where the center $(x_0, y_0)$ and the radius $r$ are known. Is there a general formula for the equation of a tangent at a point $(a, b)$ on the circle, with $a, b$ known?
Idea
Since the point $(a, b)$ is on the circle, we must have
$$
(x – a)^2 + (y – b)^2 = r^2
$$
and since the tangent will pass through $(a, b)$ we must have
$$
\frac{y – b}{x – a} = m
$$
How do you incorporate the fact that the radius is perpendicular to the tangent?
Following the comments, I think then one notices that the radius is perpendicular to the tangent. The line through the center $(x_0, y_0)$ and the point $(a, b)$ has slope
$$
m' = \frac{b – y_0}{a – x_0}
$$
Since this line must be perpendicular to the tangent, we must have that
$$
m'm = -1
$$
Therefore we can already find the slope as
$$
m = -\frac{1}{m'} = – \frac{a – x_0}{b – y_0}
$$
Now we just need to find the intercept. Plugging $(a, b)$ into $y = mx + q$ with the $m$ that we've just found we find
$$
b = ma + q
$$
and so
$$
b + a\left(\frac{a – x_0}{b – y_0}\right)
$$
The final equation is
$$
y = -\frac{a – x_0}{b – y_0} x + b + a\left(\frac{a – x_0}{b – y_0}\right)
$$
Best Answer
$$y - b = -\frac{a-x_0}{b-y_0}(x-a)$$
(Tangents are perpendicular to radii, and then just use the standard equation of a line of known gradient through a point)