The given equation of circle is, $x^2 + y^2$ – 8x + 10y – 12 = 0 ∴ $(x^2 - 8x) + (y^2 + 10y)$ = 12 ⇒ [$x^2$ – 8x + $(4)^2$] + [$y^2$ + 10y + $(5)^2$] = 12 + $(4)^2 + (5)^2$ ⇒ $(x - 4)^2 + (y + 5)^2$ = 12 + 16 + 25 ⇒ $(x - 4)^2 + (y + 5)^2$ = 53 ⇒ $(x - 4)^2 + (y + 5)^2$ = $(\sqrt{53})^2$ Comparing it with $(x - h)^2 + (y - k)^2$ = $r^2$, we have h = 4, k = –5 and r = $\sqrt{53}$. Thus co-ordinates of the centre are (4, –5) and radius is $\sqrt{53}$.