For #r#, use the formula #r=sqrt(x^2+y^2)=sqrt(5^2+5^2)=sqrt(50)=sqrt(25)sqrt(2)=5sqrt(2)#.
For #theta#, note that the point #(x,y)=(-5,-5)# is in the third quadrant. Hence, #theta=tan^{-1}(y/x)+pi=tan^{-1}(1)+pi=pi/4+pi=(5pi)/4# (adding or subtracting any whole number multiple of #pi# to this also works for #theta#).