The equation of circle is $(x-h)^2 + (y-k)^2$ = $r^2$ ....(i) Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2) ∴ radius of circle = $\sqrt{(4-2)^2+(5-2)^2}$ = $\sqrt{4+9}$ = $\sqrt{13}$ Now the required equation of circle is $(x-2)^2 + (y-2)^2$ = $(\sqrt{13})^2$ ⇒ $x^2$ + 4 – 4x + $y^2$ + 4 – 4y = 13 ⇒ $x^2$ + $y^2$ – 4x – 4y – 5 = 0.