Find point of tangency.

Question

Suppose I have a circle with radius r, with its center at (cx, cy), and a tangent line segment with the first point at (ax, ay), the second point at (bx, by). Knowing all the information above, how do I find the coordinates (x, y) of the point of tangency?

Here's an image just to be clearer

Answer 1

The straightforward approach is to work out the equations of the line and circle and solve the resulting system, but it’s possible to reduce this problem to finding the intersection of a pair of lines, which I expect you know how to do.

A tangent line is perpendicular to the radius of the circle at the point of tangency. An equation of the line through the two points (which I hope you can find for yourself) is $$(a_y-b_y)x-(a_x-b_x)y+(a_xb_y-a_yb_x)=0.$$ The perpendicular through the circle’s center is $$(b_x-a_x)x-(a_y-b_y)y= (b_x-a_x)c_x-(a_y-b_y)c_y.$$ Solve this system of equations.

Another way to reduce this problem to the intersection of two lines is to use the polars of the two points. The polar of an exterior point is the chord of contact of the tangents through that point; the two polars intersect at the sought-after point of tangency. You could also intersect either polar with the tangent line itself, but the equations of the polars are obtained by a mechanical substitution into the equation of the circle, which saves you having to compute an equation of the tangent line. Specifically, if the equation of the circle is $x^2+y^2-2c_xx-2c_yy+c_x^2+c_y^2-r^2=0$, then the polar of the point $(a_x,a_y)$ is $$a_xx+a_yy-c_x(x+a_x)-c_y(y+a_y)+c_x^2+c_y^2-r^2=0$$ and similarly for $(b_x,b_y)$.

As a third alternative, you can use the fact the tangent at a point on the circle is the polar of that point. If you compare the coefficients of the equation of the polar at the unknown point $(x_0,y_0)$ to the equation of the tangent from above (they must be in the same proportions), you’ll get a system of linear equations in the coordinates of the point of tangency and the unknown constant of proportionality. You can use a cross product to avoid introducing another unknown, but I don’t think that really saves any work here.

Answer 2

First find the equation of the tangent line, which is $y-a_y=m(x-a_x)$ where $m=\frac{a_y-b_y}{a_x-b_x}$. Simply let $y$ to be the subject,$$y=m(x-a_x)+a_y...(1)$$

Then find the equation of the circle which is $(x-c_x)^2+(y-c_y)^2=r^2$.

Substitute $(1)$ into the equation of the circle and expand all term such that it will be a quadratic equation in $x$. Solve for $x$, I belive it will have $2$ equal roots. Calculate $y$ by resubstituting the value of $x$ into $(1)$.

Note : There are also another method using derivatives, if you have learnt it.