Find two vectors from one

Question

I've made a quick sketch of my problem so it's easier to understand. My goal here is to find vector a.

I start out with only knowing the length and direction of the vector c, the hypotenuse of the triangle. I also know that the angle C is 90 degrees the radius of the circle, b, and that a is tangent to the circle.

By using Pythagoras I can find the length of a, but I don't know how to find vector a.

I probably could make a circle around the point at B with a radius of 'a', and the find the intersection between the two circles.

Are there any easier solutions?

Quick Sketch

Answer 1

Assume that $A$ is the origin, so the equation of the circle is $x^2+y^2=b^2$.

Since the vector $\underline{c}$ is known, assume we have the coordinates of $B(p,q)$

A line with gradient $m$ passing through $B$ is $$y=mx-mp+q$$

Solving this simultaneously with the circle, and applying the condition that the resulting quadratic equation in $x$ must have double roots because of the tangent point leads to, after some algebra, to the following quadratic equation in $m$: $$m^2(p^2-b^2)-2mpq+q^2-b^2=0$$

From this you can extract the two possible values of $m$, namely $$\frac{pq\pm\sqrt{b^2p^2+b^2q^2-b^4}}{p^2-b^2}$$

Answer 2

You know the lengths of $\vec{a}$ and $\vec{b}$ (which I'm going to call $a$ and $b$).

Let $$ t = atan2(b, a) $$ Then the angle of vector $\vec{a}$ is the angle of $\vec{c}$, minus $t$.

(angles measured counterclockwise from the $x$-axis).