The equation of the circle having normal at $(3, 3)$ as $y = x$ and passing through $(2, 2)$ is:

${ x }^{ 2 }+y^{ 2 }+2gx+2fy+c=0\\ 2x+2y\cfrac { dy }{ dx } +2g+2f\cfrac { dy }{ dx } =0.\\ \cfrac { dy }{ dx } =\cfrac { -y-x }{ y+f. } \\ \cfrac { dy }{ dx } \quad at\quad (3,3).\\ \Rightarrow \cfrac { dy }{ dx } =\cfrac { -y-3 }{ 3+f } \\ Slope\quad of\quad normal\quad at\quad this\quad point\quad =1\\ \Rightarrow -\cfrac { (g+3) }{ f+3 } (1)\quad =-1\\ \Rightarrow g+3=f+3.\\ \Rightarrow g=f.\\ \Rightarrow { x }^{ 2 }+y^{ 2 }+2gx+2fy+c=0\\ Circle\quad passes\quad through\quad (3,3)\& (2,2).\\ \Rightarrow g+g+bg+bg+c=0\\ \Rightarrow 12g=-18-c.\\ \Rightarrow 4+4+4g+4g+c=0.\\ 8g=-8-c.\\ -12g=-18-c\\ -4g=10\\ g=\cfrac { -5 }{ 2 } =f.\\ C=-8(g+1)\\ =-8(\cfrac { -5 }{ 2 } +1)\\ =-8\times \cfrac { -3 }{ 2 } \\ =+12\\ \therefore \quad Equation\quad :\\ { x }^{ 2 }+y^{ 2 }-5x-5y+12=0.\\ Ans.$