Equation of a circle passing through two points and subtending an angle $\theta$ at the circumference.

Question

I am trying to derive the general form of the equation of the circle given two points $(x_1,y_1)$, $(x_2,y_2)$and the angle $\theta$ subtended by the chord joining the two points.

So, $\theta$ is basically the angle between two lines $\implies \tan \theta = |\dfrac{\dfrac{y-y_1}{x-x_1}-\dfrac{y-y_2}{x-x_2}}{1+( \dfrac{y-y_1}{x-x_1}\dfrac{y-y_2}{x-x_2})}| \\ \implies(x-x_1) (x-x_2)+(y-y_1)(y-y_2)= \pm \cot\theta[(y-y_1)(x-x_2)-(x-x_1)(y-y_2)]$

But this is the equation of the major arc right? How do I derive the equation of the circle from it?

Answer 1

You're done if you find the center of the circle (since you will immediately know the radius as well). It's the place where the two radii from these two points meet. There are two possibilities for this center, one on either side of the line joining the two points. Let $\alpha$ be the angle that the line connecting $(x_1,y_1)$ to $(x_2,y_2)$ forms with the x-axis. So $\tan \alpha = \frac{y_2-y_1}{x_2-x_1}$ Consider the triangle formed by the center and the 2 given points. The angles of this triangle at each of the two points is $\frac{\pi}{2}-\frac{\theta}{2}$.

Case 1: the slope of the line to the center from $(x_1,y_1)$ is $m_1=\tan(\frac{\pi}{2}-\frac{\theta}{2} + \alpha)$. the slope of the line to the center from $(x_2,y_2)$ is $m_2=\tan (\frac{\pi}{2}+\frac{\theta}{2} + \alpha)$. Knowing $m_1,x_1,y_1$ and $m_2,x_2,y_2$, you can solve for the center.

Case 1: the slope of the line to the center from $(x_1,y_1)$ is $m_1=\tan(-\frac{\pi}{2}+\frac{\theta}{2} + \alpha)$. The slope of the line to the center from $(x_2,y_2)$ is $m_2=\tan (\frac{3\pi}{2}-\frac{\theta}{2} + \alpha)$. Knowing $m_1,x_1,y_1$ and $m_2,x_2,y_2$, you can solve for the center.

Answer 2

If the radius of the circle is $r$, then the distance between the two points is $2r\sin\frac\theta2$, therefore $$r^2=\frac14\left((x_1-x_2)^2+(y_1-y_2)^2\right)\csc^2\frac\theta2.$$ In general, there are two possibilities for the circle’s center, one on either side of the line through the two points. There are a few straightforward ways to find these centers. One is to compute the intersection of the two circles with radius $r$ centered at the two given points. Since these centers lie on the perpendicular bisector of the given points, you can simplify this a bit by instead computing the intersection of one of these circles and this line, which is given by the equation $$2(x_2-x_1)x+2(y_2-y_1)y=(x_2^2-x_1^2)+(y_2^2-y_1^2).$$ However, you can also compute the centers directly. From the midpoint, go a distance of $r\cos\frac\theta2$ along the perpendicular bisector, i.e., in a direction perpendicular to the chord joining the two points. This gives us $$x_c = \frac12(x_1+x_2)\pm{y_1-y_2\over\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}r\cos\frac\theta2 \\ y_c = \frac12(y_1+y_2)\pm{x_2-x_1\over\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}r\cos\frac\theta2$$ as the coordinates of the circle centers. I’ll leave constructing and simplifying the resulting circle equations to you.