What is the equation of a tangent to a circle at a point?

Question

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 $$ y = -\frac{a - x_0}{b - y_0} x + b + a\left(\frac{a - x_0}{b - y_0}\right) $$

Answer 1

$$y - b = -\frac{a-x_0}{b-y_0}(x-a)$$

(Tangents are perpendicular to radii, and then just use the standard equation of a line of known gradient through a point)

Answer 2

HINT:

The tangent should pass through the circumference point (a, b) and since its perpendicular to the radius, it should have a slope negative reciprocal of the slope of the radius vector

Answer 3

(A bit different take on this from the other answers...)

Yes.

$$(x-a)^2+(y-b)^2+r^2=(x-x_0)^2+(y-y_0)^2$$

This formula comes from Pythagoras' theorem applied to the right-angled triangle with corners $(x,y)$, $(x_0, y_0)$ and $(a,b)$. It is a formula for a line because $x^2$ and $y^2$ will cancel when you expand LHS and RHS.

Answer 4

Label $(x_0, y_0)$ as $O$, $(a,b)$ as $T$ and let $P$ be an arbitrary point on the tangent line at $(x,y)$.

Then the dot product $OT.TP = 0$ because the vectors need to be perpendicular, which calculates out as

$(a-x_0)(x-a) + (b-y_0)(y-b) = 0$
$(y-b) = \dfrac{(a-x_0)}{(b-y_0)}(a-x)$