Find the equation of a circle with centre $(2,2)$ and passes through the point $(4,5)$.

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 point $$(2, 2)$$ and $$(4, 5)$$
$$\displaystyle

\therefore r = \sqrt{\left ( 2 - 4 \right )^{2}+\left ( 2 - 5 \right

)^{2}}= \sqrt{\left ( -2 \right )^{2}+\left ( -3 \right

)^{2}=}$$$$\displaystyle \sqrt{4 + 9} =\sqrt{13}$$
Thus the equation of the circle is 
$$\displaystyle \left ( x - h \right )^{2} +\left ( y - k \right )^{2} = r^{2}$$
$$\displaystyle \left ( x - 2 \right )^{2}+\left ( y - 2 \right )^{2}=\left ( \sqrt{13} \right )^{2}$$
$$\displaystyle x^{2}-4x + 4 + y^{2}- 4y + 4 = 13$$
$$\displaystyle x^{2}+y^{2} - 4x - 4y - 5 = 0$$