I have a triangle circumscribed within a circle. One vertex is the exact center of the circle. To solve the question, I need to figure out the exact coordinates of the center point of the circle.




• $(x, y)$ and $(a, b)$ lie on the circumference of the circle.

• $(h, k)$ is the center of the circle.

• $(a, b)$ is known. $(x, y)$ and $(h, k)$ are unknown.

• $r$ is the radius of the circle. $r$ is known.

• $θ$ is the degrees of rotation that $(a, b)$ has been rotated clockwise from $(x, y)$.

• $θ$ is known.

• $u$ is unknown but can be calculated because the triangle is isosceles (which means the two remaining angles are congruent. Therefore $u / sinθ = r / sin((180 - θ) / 2)$. where $((180 - θ) / 2)$ is the formula used to get the degrees of either missing angle.

• (x, y) can be interpreted as (h, k + r).

Is it actually possible to find the center of the circle using the given information? I realize this isn't a standard geometry question, but real life applications of math rarely follow that format.

  • $\begingroup$ Strictly speaking, your triangle is not inscribed in the circle. It is within. $\endgroup$
    – Jean Marie
    Mar 16 '17 at 18:01

The answer is yes and $(h,k) = (a-r \sin \theta, b-r \cos \theta)$.

enter image description here

According to your description, the side of the triangle opposite point $(a,b)$ is parallel to the y-axis. So, start by finding the height of the triangle from point $(a,b)$, which is $h_1=r \sin \theta$. Then find the length $dy = r \cos \theta$. Works for $\theta \lt 180^\circ$.


Just noticed I used $h$ in two different ways. Changed the height to $h_1$ in both the figure and the text.

  • $\begingroup$ This was the perfect response. Worked like a charm, and I don't think I would have figured it out if you hadn't posted this. $\endgroup$ Mar 15 '17 at 20:05
  • $\begingroup$ Glad to help! :-) $\endgroup$
    – Jens
    Mar 15 '17 at 20:13

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