Given a circle $x^2+y^2=r^2$ and a point P(x1,y1) outside of the circle. I can draw two tangents from P to the circle. I will call A and B the points where the tangents cross the circle. How can I show that the line that passes through A and B is defined by $x.x_1+y.y_1=r^2$?
The slope of the line is $-x_1/y_1$ because that line is perpendicular to the line that goes from (0,0) to (x1,y1).
How can I show that the intercept is $r^2/y_1$?
Suppose that we know how to find the equation of the tangent at a point on the circle.
Here is another approach to show that $x_1 x + y_1 y=r^2 $ is the equation of $AB$.
Note that the equation of the tangent at $A(\alpha, \beta)$ is: $$x \alpha + y \beta=r^2$$ Similarly the equation of the tangent at $B(\gamma, \delta)$ is: $$x \gamma + y \delta = r^2$$
Since $P(x_1, y_1)$ lies on both tangents, therefore
$$x_1 \alpha + y_1 \beta=r^2 \tag{1}$$ and $$x_1 \gamma + y_1 \delta = r^2 \tag{2}$$
Now consider the straight line equation $$x_1 x + y_1 y=r^2 \tag{*}$$
Notice that $A(\alpha, \beta)$ satisfies $(*)$ because of $(1)$.
Similarly $B(\gamma, \delta)$ satisfies $(*)$ because of $(2)$.
Hence $$x_1 x + y_1 y=r^2 $$ is the equation of the straight line passing through $A$ and $B$.
