Q.Find the centre and radius of the circle $x^2+y^2-6x+4y-12=0$.

A circle is the locus of points at a fixed distance (the radius $r$) from a fixed centre $(h,k)$; applying the distance formula to this definition gives the standard equation $(x-h)^2+(y-k)^2=r^2$, which reduces to $x^2+y^2=r^2$ when the centre is the origin. Expanding the standard form produces the general equation $x^2+y^2+2gx+2fy+c=0$, from which the centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ can always be recovered, either by direct substitution into these formulas or by completing the square. The sign of $g^2+f^2-c$ determines whether the equation represents a genuine circle (positive), degenerates to a single point (zero, matching the point degenerate conic), or has no real solutions at all (negative). Three independent pieces of data -- a centre and a radius, a centre and one point, or three points on the circle -- are always enough to pin the equation down uniquely.