Prove the geodesic on 2-sphere is the great circle

I want to use the Killing vector fields to prove the geodesic on the sphere is the great circle. First of all, the given metric is $$ds^2=d\theta^{2}+\sin^2\theta d\phi^{2},$$ where I set the radius to be $1$. By Killing equation, $$\nabla_\mu K_\nu+\nabla_\nu K_\mu =0,$$ and some computation, I have the following three Killing vectors: \begin{align*} K_1 &= \partial_{\phi} \\ K_2 &= \cos\phi \, \partial_{\theta} - \cot\theta \sin\phi \partial_{\phi} \\ K_3 &= -\sin\phi \partial_{\theta} - \cot\theta \cos\phi \partial_{\phi} \end{align*} I want to use the fact that $$\frac{d}{d\lambda}\left\{K_\mu \frac{dx^{\mu}}{d\lambda}\right\}=0,$$ where $x$ is the curve and $\frac{dx^{\mu}}{d\lambda}$ is the tangent vector. However, as everyone knows, the geodesic is the great circle. Therefore, I thought the final equation would be of the form, $$aX+bY=0,$$ where $a$ and $b$ are some constant, while $X$ and $Y$ are positions on the sphere. The equation passes the origin. But I can't see how to get the desired results, please give me some help.

This should probably be a comment, but I do not yet have the reputation to make a comment. However, I believe your final form of the solution is incorrect. A great circle does not pass through the origin of a sphere: the center of the circle does, but the center of the circle does not belong to the curve. If $$X$$ and $$Y$$ are positions on the sphere, then you should have the final form
$$X^2+Y^2=R^2$$
where $$R$$ is the radius of the sphere. However, I do not think this will tell you much because $$X$$ and $$Y$$ can be arbitrary, so long as they belong to the sphere: it is not necessarily true that three points on the sphere can be connected by a great circle. I suggest working in spherical coordinates, and from there, you expect that at the equator of the sphere, a geodesic follows the equator.