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.:)
Best Answer
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.