The equation of the circle which passes through the point (4, 5) and its centre at (2, 2) is

The equation of the circle is given by $$\left(x-a\right)^2+\left(y-b\right)^2=r^2$$, where $$(a,b)$$ represents the center of the circle and $$r$$ represents the radius of the circle.

We know the centre of the circle is at (2,2) 

$$\left(x-2\right)^2+\left(y-2\right)^2=r^2$$

Also, the circle is passing through the point (4,5). Thus, we can substitute x = 4 and y = 5 -

$$\left(4-2\right)^2+\left(5-2\right)^2=r^2$$

$$r^2=4+9$$

$$r^2=13$$

Thus, the equation of the circle can be given as $$\left(x-a\right)^2+\left(y-b\right)^2=13$$