Find vector tangent to circle through a point

Question

C2 and P1 are known points. Radius of circle C2 is also constant and known. I am trying to find point T to eventually construct line p1-t, which is tangent to circle c2.

I have made an attempt involving bisecting c2-p1 at M, and performing trigonometric operations to find measure of angle TMC2. Then, I found the difference in x and y coordinates from m.x and m.y using a polar form to cartesian (polar: angle tmc2 and length of tm = c2m = 1/2 c2P). After this, I normalized vectors c2P and added the difference in x, normalized the vector perpendicular to c2p and added the difference in y.

This approach seems very convoluted for such a simple task. Are there any basic vector operations to save time and processing power?

Thank you

Answer 1

It's pretty easy with some trigonometry.

Here's a hand drawn image of your problem:

OTP angle is a right angle.

Calculating l is done with the pythagorean theorem:

l = sqrt(d^2 - r^2)

You first want to calculate alpha. This is where the trigonometry part comes in.

tan(alpha) = r / l
alpha = atan2(r, l)

After you have alpha, you need to calculate the angle of PT relative to the coordinate system (let' call it beta). This is done by calculating the angle of PO (let's call it gamma) then subtracting alpha from it:

vector := O - P // vector pointing from P to O
gamma = atan2(vector.y, vector.x)
beta := gamma - alpha

Then finally you convert it to a directional vector with a length l and you add P to it

x := cos(beta) * l + P.x
y := sin(beta) * l + P.y

Answer 2

You can solve this without trigonometric functions, too.

Since the vector u = T - C is perpendicular to the vector v = T - P, the dot product u . v = 0. You also know the lengths of 2 sides of a right triangle (the radius r and ||C - P||).

This allows you to write 2 equations in the 2 unknown elements of T, which you can then solve:

(T-C) . (T-P) = 0
        |T-P| = |C-P| - r^2

where |X| is the squared length of the vector X.

But this might be annoying to solve, so you can further exploit that in 2D, the vector [x,y] is perpendicular to the vector [-y,x]. And so we can instead write the equation:

T = P + u' t = C + u r

where u' is [-uy,ux] and t = ||T-P|| = sqrt(|C-P| - r^2)--which we have to solve for u. This gives us:

ux = (r * (Px - Cx) - t * (Py - Cy))/(r^2 + t^2)
uy = (r * (Py - Cy) + t * (Px - Cx))/(r^2 + t^2)

and therefore:

Tx = Cx + ux * r
Ty = Cy + uy * r

There is an unfortunate caveat--one other vector is perpendicular to [x,y] at [y,-x], which is a clockwise rotation instead of counterclockwise. There are simple tests you can use to decide which to go with, but one or the other might suit you in practice.

This all involves only one sqrt call and so it may be less computationally expensive than Baliant's answer.