A right angled isosceles triangle is incribed in the circle x2+y2-4x-2y-4=0 then length of its side is

A) square root of 2 B) 2*square root of 2 C) 3*square root of 2 D) 4*square root of 2

The given circle is, 
x² +y² -4x-2y-4 = 0 
Now comparing this equation of circle with standard equation of circle, 
x² +y² +2gx + 2fy +c = 0 ​we get, 

                      
g = -2 , f = -1 c = -4 so the center of the circle will be (-g, -f) or (2, 1) and radius can be calculated as, 

$$\begin{gathered} r =\sqrt{g^2+f^2-c} \\ r = \sqrt{2^2+1^2-(-4)} \\ =\sqrt{9} = 3 \end{gathered}$$
Since given triangle is a right angle isosceles inscribed in the circle so , 
BC will be diameter as shown in the figure and AB= AC , 
Now BC= 2r = 6 unit
Now in right angle triangle BAC , 
$$\begin{gathered} \Rightarrow A{B}^2+A{C}^2=B{C}^2 \\ \Rightarrow A{B}^2+A{B}^2=6^2 \\ \Rightarrow 2A{B}^2=36 \\ AB=\sqrt{18}=3\sqrt{2} unit \end{gathered}$$