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

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?

• This is gives the equation of circle only! Note $\cot(\pi-x) = -\cot(x)$. – samjoe Sep 28 '17 at 9:07
• If you use $\pm$, then you will get two circles. – samjoe Sep 28 '17 at 9:14

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.

• I try to avoid working with slopes when there’s a possibility of vertical lines since those end up requiring special cases. – amd Sep 28 '17 at 19:56
• You cannot avoid that. Even in the expression you have written as part of the question, there is $\tan \theta$, $\cot \theta$ etc which can be infinity. I I wouldn't shrink from it because of the $\infty$. Many trigonometric identities have $\infty = \infty$ in special cases. For instance $\tan (a+b) = \frac{\tan a + \tan b}{1- \tan a \tan b}$ when $a+b=\frac{\pi}{2}$ – Mathemagical Sep 29 '17 at 2:13
• Infinities and other special cases are entirely avoidable in this problem. – amd Sep 29 '17 at 7:12
• What if the points are (1,0) and (0,1) and the angle $\theta=\frac{\pi}{2}$? This is just the unit circle, so the equation of the circle is quite well defined. But of course, one of the lines to the center is vertical with infinite slope. That is my point - don't be scared of infinity, avoiding it at all costs. Slope being infinity in an intermediate step doesn't mean that there will be something wrong with the final circle equation. – Mathemagical Sep 29 '17 at 7:20
• It’s just as easy, if not easier, to work with the point-normal form of the equation of a line in this problem. All of the necessary computations can be done without ever introducing slope and its concomitant infinities. – amd Sep 29 '17 at 7:47

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.