# Given the equation of a circle and a point on a tangent to the circle, determine the point of contact of the tangent to the circle

I was looking through my old high school mathematics textbook and stumbled across a question with the diagram below. The radius of the circle is root 14 and BC is a tangent to the circle at point C. The question was determining the length of BC which I did not find difficult at all. The question stopped there, but I wanted to find the co-ordinates of point C. I have tried using to get a system of linear equations to solve simultaneously, but I end up with too many unknowns. I do not know how to proceed. Is there a general argument which can be used for a problem like this and if so could I have a hint so that I may attempt to solve this myself.

• You can calculate the distance $BC$, since it is on a right triangle. You know the leg $AC$(which is the radius) and the hypothenuse(which is $AB$). Then, you can make a quadratic equation on $AC$ and on $BC$. It shall give a system of two equations and two unknowns($C_x$ and $C_y$) – Cristhian Grundmann Jul 24 '18 at 21:43
• The radius looks closer to 4 than 14 (if the radius was 14, B would be inside the circle and no solution would exist) – WW1 Jul 24 '18 at 21:50
• Ah my bad it is the root of 14 – Herb Jul 24 '18 at 21:51

A system of linear equations will give you either zero, one or an infinite number of solutions. There are two tangent lines through $B$, hence also two points of tangency, so it’s not at all surprising that you’ve not been able to come up with a system of linear equations for the coordinates of $C$.
You have the circle’s radius, and you say that you were able to work out the distance $r=BC$. The point $C$ is one of the two intersections of the given circle with a circle of radius $r$ centered at $B$. This gives you a system of second-degree equations in $x$ and $y$ to solve. If you subtract one of these equations from the other, you’ll get a linear equation in $x$ and $y$ (in fact, the equation of the line through the two intersection points). Solve for one of the variables in terms of the other and back-substitute into one of the circle equations to get a quadratic in one variable.
Since you have the length of BC and also know that the length of AC is the radius of the circle, you can compute the angle between AC and AB as $\alpha = \tan^{-1} \frac{|BC|}{r}$.
You should be able to form an equation for the line passing through points A and B. Then you can also find the equation for the line passing through A that makes the angle $\alpha$ with the line AB. This can be done by considering the known unit vector $r_1$ in the direction of the line AB and the unknown unit vector $r_2$ in the direction of AC. Their dot product must satisfy $(r_1,r_2) = r_{1,x}r_{2,x} + r_{1,y}r_{2,y} =||r_1||||r_2||\cos \alpha = \cos \alpha$, and also it must be that $||r_2||^2 = r_{2,x}^2 + r_{2,y}^2 = 1.$ Thus you have two equations for the two unknown components of the unit vector $r_2$. After solving $r_2$ you can get the equation for the line AC by requiring that the line passes through A.
Once you have the equation for the line AC, you can find point C a distance $r$ away from point A on this line.
• I wouldn’t bother computing $\alpha$ explicitly since you really want its cosine, which you can compute directly from the tangent. – amd Jul 25 '18 at 0:35