How do i describe the circle resulting from the intersection of the plane $ax+by+cz+d=0$ and the sphere centred at $O(X_0,Y_0,Z_0)$ with radius $R$
I found that the plane-sphere distance $\rho$ is less than the radius of the sphere so a circle is formed I also found from my diagram that the radius $r$ of my circle follows $r = \sqrt{R^2-{\rho}^2}$ But how do I go about finding the coordinates of the centre
Let $\vec{n}$ be normal to the plane, $O'$ the centre of the circle I thought of saying $\vec{n} \times \vec{OO'} = \vec{0}$ Since $\vec{OO'}$ is parallel to $\vec{n}$
Thanks in advance
The centre of your circle is the intersection of the plane with the line that passes through the centre of the sphere and is perpendicular to the plane.