Find the equation of a circle with centre (2, 2)

The centre of the circle is given as (h, k) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).

$\therefore r=\sqrt{(2-4)^{2}+(2-5)^{2}}=\sqrt{(-2)^{2}+(-3)^{2}}=\sqrt{4+9}=\sqrt{13}$

Thus, the equation of the circle is

$(x-h)^{2}+(y-k)^{2}=r^{2}$

$(x-2)^{2}+(y-2)^{2}=(\sqrt{13})^{2}$

$x^{2}-4 x+4+y^{2}-4 y+4=13$

$x^{2}+y^{2}-4 x-4 y-5=0$