Given a point outside of an arc, how can one find the point on the arc which extends to that point?
For example, the radius of the circle (R) is 10cm, it's center point is [0,0].
The origin (o) of the line (8) is at [-3, 10]
How can we find the point (p) (p8) were the tangent at that point continues to the origin of the line?
A brute force solution would not be acceptable.
Best Answer
If i understood your question correctly, you want to find the point at which the tangent of the circle that passes through [-3, 10] touches the circle. Using the fact that a radius to that point will be perpendicular to the tangent, we can find that point quite easily. Lets build the triangle with vertices in [-3, 10], [0, 0] and the point we are trying to find. It will be a right triangle, so we can use the Pythagoras theorem. We know the length of one of the catheti is 10 (radius), and the hypotenuse is $\sqrt{3^2 + 10^2} = \sqrt{109}$. The length of the second cathetus will therefore be $\sqrt{109 - 10^2} = 3$. Lets build a circle with this radius and the center in [-3, 10]. The point of intersection of two circles will be the desired point.