I am facing trouble finding out the equation of the circle which passes through two points ($x_1,y_1$) and ($x_2,y_2$) and the chord joining ($x_1,y_1$) and ($x_2,y_2$) making an angle $\theta$ in the major segment of the circle.

I thought of transforming the equation to polar coordinates and taking the line joining ($x_1,y_1$) and ($x_2,y_2$) as the initial line and ($x_1,y_1$) as the origin. The equation of the circle in the new system would be then $$r^2-2ar\sin(\theta-\theta')=0$$ But I am facing difficultly restoring it to the origin coordinate axes and hence converting it in cartesian form.

Any help in this regard would be highly appreciated.Thanks.:)


2 Answers 2


Let your two known points be $A$ and $B$. Let the center of the circle (which you have to find) be $P$. You are given that $A$ and $B$ are both on the circle and you are given the angle $\angle APB = \theta$. (Or perhaps it is $\angle APB = 2\pi - \theta$, or perhaps $\angle APB = 2\theta$, depending on what is meant by "making an angle $\theta$ in the major segment of the circle." In any case, in this answer I'll take $\angle APB$ as known and will not use the symbol $\theta$.)

Let $M$ be the midpoint of segment $\overline{AB}$. Then $\triangle AMP$ and $\triangle BMP$ are congruent right triangles. The hypotenuse of each triangle, $PA$ or $PB$, equals the radius of the circle, one leg is equal to $\frac12(AB)$, and the other leg is equal to $MP$. Note that since $A$ and $B$ are known, the length $AB$ is easily found.

We also know that $\angle APM = \frac12 \angle APB$. So we have a right triangle with one leg ($\frac12(AB)$) and the angle opposite that leg ($\frac12 \angle APB$) are known. Therefore we can find the length of the other leg of the triangle by using trigonometry.

So now we have the distance $MP$. We also know $P$ must be on the perpendicular bisector of $\overline{AB}$. With that information we can find the coordinates of $P$; there are two possible results, depending on which direction you go along the perpendicular bisector. The easiest way to find one of the possible locations of $P$ may be to take $AB$ and $MP$ as the hypotenuses of two right triangles whose legs are parallel to the $x$ and $y$ axes; the two triangles are similar, so the legs of one are easily computed when you known its hypotenuse $MP$ and all three sides of the other triangle.

At this point in the procedure you have the $(x,y)$ coordinates of the center of the circle, and there are several ways you could find its radius. Writing the equation of the circle is simply a matter of plugging this information into the well-known formula.

  • $\begingroup$ Please consider adding diagrams to this answer, $\endgroup$
    – Archer
    Sep 28, 2017 at 7:07

Let $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x,y)$ be a point on the major arc $AB$.

By simple geometry of circle,

\begin{align*} \angle AOB &= 2\angle ACB \\ 2\pi-\theta &= 2\angle ACB \\ \end{align*}

Applying cosine law in $\Delta ABC$, \begin{align*} \cos \angle ACB &= \frac{a^2+b^2-c^2}{2ab} \\ -\sin \frac{\theta}{2} &= \frac{(x-x_1)(x-x_2)+(y-y_1)(y-y_2)} {\sqrt{(x-x_1)^2+(y-y_1)^2}\sqrt{(x-x_2)^2+(y-y_2)^2}} \end{align*}

which is the equation of major arc $ACB$.

Squaring both sides gives two circles.

For example, $\theta=270^{\circ}$, $A=(1,1)$, $B=(-1,-1)$ we have

\begin{align*} x^4+2x^2y^2+y^4-8x^2+8xy-8y^2+4 &= 0 \\ (x^2+y^2+2x-2y-2)(x^2+y^2-2x+2y-2) &= 0 \end{align*}

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  • $\begingroup$ How is angle AOB =$2\pi-\theta$ $\endgroup$
    – Archer
    Sep 28, 2017 at 7:09
  • $\begingroup$ $\angle AOB$ is the the angle of minor arc while $\theta$ is angle of minor arc. $\endgroup$ Oct 3, 2017 at 17:37

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