Finding the point of contact for circle tangent without trigonometry

Question

Given that we know $(x_1, y_1)$ and a circle with center (0,0) and radius R.

What is the points of contact the tangents? ( $(x_2,y_2)$ and $(x_3,y_3)$ respectively, in the Figure. )

I am think I am able to set up two equations using Pythagaros:

$x_2^2+y_2^2=R^2$

$(x_2-x_1)^2 +(y_2-y_1)^2 = x_1^2+y_1^2-R^2$

However I am not able to solve this any further.

What I would like is to get a formula on the form (as simple as possible):

$x_2/x_3$ =

$y_2/y_3$ =

As a note I think it would be fairly easy to express it using trigonometry (using sine, and tangens), but I would like to express it without using angles.

Problem Description

Answer 1

Following your idea we have that the distance $P_1=(x_1,y_1)$ form the center is

• $D_0^2=x_1^2+y_1^2$

and thus the distance $P_1=(x_1,y_1)$ form $P_2=(x_2,y_2)$ is

• $D^2=(x-x_1)^2+(y-y_1)^2=D_0^2-R^2$

therefore the solution is given by the system

• $(x-x_1)^2+(y-y_1)^2=D_0^2-R^2$
• $x^2+y^2=R^2$

Answer 2

Calling

$$ L\to p=p_1+\lambda \vec v\\ C\to||p-p_0|| = R $$

then $C \cap L\to ||p_1-p_0+\lambda\vec v||^2 = R^2$ or

$$ ||p_1-p_0||^2+2\lambda <p_1-p_0,\vec v> +\lambda^2||\vec v||^2 = R^2 $$

now solving for $lambda$

$$ \lambda = \frac{-2<p_1-p_0,\vec v > \pm \sqrt{4< p_1-p_0,\vec v>^2-4||\vec v||^2(||p_1-p_0||^2-R^2)}}{2||\vec v||^2} $$

but we need tangency hence

$$ 4< p_1-p_0,\vec v>^2-4||\vec v||^2(||p_1-p_0||^2-R^2)=0 $$

now calling $m = \frac{v_y}{v_x}$ and considering $p_0 = (0,0)$

$$ < p_1,\vec v>^2-||\vec v||^2(||p_1||^2-R^2)=0 $$

or

$$ (m x_1+y_1)^2-(m^2+1)(x_1^2+y_1^2-R^2) = 0 $$

or

$$ m =\frac{- x_1 y_1 \pm R\sqrt{x_1^2+y_1^2-R^2}}{R^2-y_1^2} = \frac{y_t-y_1}{x_t-x_1} $$

where $(x_t, y_t)$ is one of the tangent points.

Answer 3

Summary of results: If $c^2 = (x_1^2+y_1^2)/R^2$ then $x=\dfrac{x_1\pm y_1\sqrt{c^2-1}}{c^2}$ and $y=\dfrac{y_1\mp x_1\sqrt{c^2-1}}{c^2}$.

Suppose $(x_2, y_2)$ is a point on the circle, so $x_2^2+y^2_2 = R^2$. The line from $(x_1, y_1)$ to $(x_2, y_2)$ is $\dfrac{y-y_1}{x-x_1} =\dfrac{y_2-y_1}{x_2-x_1} $ or $(y-y_1)dx = (x-x_1)dy$ where $dx = x_2-x_1$ and $dy = y_2-y_1$.

The slope of this line is $\dfrac{dy}{dx}$.

The slope of the line from $(0, 0)$ to $(x_2, y_2)$ is $\dfrac{y_2}{x_2}$.

For the line to be a tangent, the product of these slopes must be $-1$, so that $-1 =\dfrac{y_2}{x_2}\dfrac{dy}{dx} $ or $(y_2-y_1)y_2 =-(x_2-x_1)x_2 $ so $y_2^2+x_2^2 =y_1y_2+x_1x_2 $.

Since $y_2^2+x_2^2 = R^2$, $R^2 =y_1y_2+x_1x_2 $.

Removing the "$_2$", the equations are $x^2+y^2 = R^2$ and $xx_1+yy_1 = R^2$. from the second, $y = (R^2-xx_1)/y_1$. Substituting in the first,

$\begin{array}\\ x^2 &=R^2-y^2\\ &=R^2-((R^2-xx_1)/y_1)^2\\ \text{or}\\ x^2y_1^2 &=y_1^2R^2-(R^2-xx_1)^2\\ &=y_1^2R^2-(R^4-2xx_1R^2+x^2x_1^2)\\ &=y_1^2R^2-R^4+2xx_1R^2-x^2x_1^2\\ \end{array} $

so $0 =x^2(y_1^2+x_1^2)-2xx_1R^2+R^4-y_1^2R^2 =dx^2-2xx_1R^2+R^4-y_1^2R^2 $ where $d=y_1^2+x_1^2$.

Solving,

$\begin{array}\\ x &=\dfrac{2x_1R^2\pm\sqrt{4x_1^2R^4-4d(R^4-y_1^2R^2)}}{2d}\\ &=\dfrac{x_1R^2\pm R\sqrt{x_1^2R^2-d(R^2-y_1^2)}}{d}\\ &=\dfrac{x_1R^2\pm R\sqrt{x_1^2R^2-dR^2+dy_1^2)}}{d}\\ &=\dfrac{x_1R^2\pm R\sqrt{(x_1^2-d)R^2+dy_1^2)}}{d}\\ &=R\dfrac{x_1R\pm\sqrt{-y_1^2R^2+dy_1^2)}}{d}\\ &=R\dfrac{x_1R\pm\sqrt{y_1^2(x_1^2+y_1^2-R^2)}}{x_1^2+y_1^2}\\ &=R\dfrac{x_1R\pm\sqrt{y_1^2(c^2R^2-R^2)}}{c^2R^2} \qquad\text{where }c^2 = (x_1^2+y_1^2)/R^2\\ &=\dfrac{x_1\pm\sqrt{y_1^2(c^2-1)}}{c^2}\\ &=\dfrac{x_1\pm y_1\sqrt{c^2-1}}{c^2}\\ \text{and}\\ y &=(R^2-xx_1)/y_1\\ &=((x_1^2+y_1^2)/c^2-(\dfrac{x_1\pm y_1\sqrt{c^2-1}}{c^2})x_1)/y_1\\ &=\dfrac{x_1^2+y_1^2-x_1^2\mp x_1y_1\sqrt{c^2-1}}{c^2y_1}\\ &=\dfrac{y_1^2\mp x_1y_1\sqrt{c^2-1}}{c^2y_1}\\ &=\dfrac{y_1\mp x_1\sqrt{c^2-1}}{c^2}\\ \end{array} $

As a check, if $(x_1, y_1)$ is on the circle so that $c^2 = 1$, then $x=x_1 $ and $y=y_1 $ which is comforting.

Note: I found dimensional analysis (making sure that the exponent of length matched) very useful in checking my algebra as I went along.

Answer 4

As a second answer, here's what trigonometry can do for you. To be completely trigish, specify the point by angle $v$ and distance $d$ so it is at $(d\cos(v), d\sin(v))$.

If the circle has radius $r$, let a point on the circle be $(r\cos(t), r\sin(t)) =(rc, rs)$.

Since the slope of the line from the center to this point is $\tan(t) =\frac{s}{c}$, the slope of the normal is $-\frac{c}{s}$.

The equation of the normal is thus $\dfrac{y-rs}{x-rc} =-\dfrac{c}{s} $ so that $0 = s(y-rs)+c(x-rc) = sy-s^2r+cx-c^2r =sy+cx-r $ or $r =sy+cx =\sin(t)d\sin(v)+\cos(t)d\cos(v) $ or $r/d =\sin(t)\sin(v)+\cos(t)\cos(v) =\cos(t-v) $ so $t-v =\arccos(\frac{r}{d}) $ and, finally $t =v+\arccos(\frac{r}{d}) $.

To get the point, let $w = \frac{r}{d}$. then $t = v+\arccos(w) $ so

$\begin{array}\\ \cos(t) &=\cos(v+\arccos(w))\\ &=\cos(v)\cos(\arccos(w))-\sin(v)\sin(\arccos(w))\\ &=\cos(v)w-\sin(v)\sqrt{1-w^2}\\ \end{array} $

and similarly for $\sin(t)$.