Q.Find the centre and radius of the circle $\left(x-\dfrac12\right)^2+\left(y+\dfrac13\right)^2=\dfrac{1}{36}$.

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 c …