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

Question

Professor PC Thomas Classes · Answer

Radius $= \sqrt{(4-2)^2+(5-2)^2} =\ \sqrt{4+9} = \sqrt{13}$. 
Equation: $(x-2)^2+(y-2)^2 = 13 \ \implies x^2-4x+4+y^2-4y+4=13 \ \implies x^2+y^2-4x-4y-5=0$.