# Finding the circumference of a spherical circle

Let $$P\in{S^2}\subset\mathbb{R^3}$$, a point on the unit sphere. Let there be a spherical circle around $$P$$ of radius $$r<\pi$$. Show that the circumference of the spherical circle is $$2\pi{\sin{r}}$$. And show that the spherical disc bounded by this spherical circle has an area of $$2\pi{(1-\cos{r})}$$.

For the first part I am aware that what needs to be understood is that a spherical circle lies on a plane in $$\mathbb{R^3}$$. Then we can consider the circle of intersection, and if we can find this radius, we can find the circumference of the intersection circle (i.e. the spherical circle).

Edit: I now have constructed a triangle, with indices of $$P$$, the origin of $$S^2$$ and some point on the spherical circle. The two radial lines have side length 1 with an angle of $$r$$. from this we can use the cosine rule to find the opposite side length and hopefully find the length of the radius of the circle in $$\mathbb{R^3}$$.

You have a good start. In the unit circle you can in fact figure out many of the things you need. This way you get the radius $$\sin(r)$$ for the circle in $$\mathbb{R}$$ to calculate the circumference.
The area can be found in some formularies. Alternatively the general formula for surfaces of revolution could be used. The function describing the unit-circle is $$f(x) = \sqrt{1-x^2}$$ and you want the area of the surface of revolution around the $$x$$-axis from $$\cos(r)$$ to $$1$$. $$A=\int_{\cos(r)}^{1} f(x) \sqrt{1+f'(x)^2} dx$$ Solving this integral results in the stated area.
• Brilliant, this has helped me so much. I'm assuming you centred $\mathbb{R^3}$ at the point at which the line OP intersects the plane I was considering. – Sam.S May 2 at 21:45