Q.Find the equation of the circle with centre at $(-3,-2)$ and radius $6$.

When a circle's centre $C(h,k)$ is not the origin, the same distance-formula argument used for the standard form gives $(x-h)^2+(y-k)^2=r^2$: for any point $P(x,y)$ on the circle, $CP=r$, and squaring gives this equation directly. Reading it in reverse, any equation of this shape can be matched against the pattern to read off the centre $(h,k)$ (the values being subtracted from $x$ and $y$) and the radius $r$ (the square root of the right-hand side) without any further computation. This form is the natural target whenever a problem gives (or lets you compute) a centre and a radius — e.g. a circle touching an axis (radius = distance from the centre to that axis), touching a given line (radius = perpendicular distance from the centre to the line), or passing through a known point (radius = distance from the centre to that point). It generalises the standard form (Section 6.1.1), which is just the special case $h=k=0$.