The centre of the circle $r^2-4 r(\cos \theta+\sin \theta)-4=0$ in cartesian coordinates is

• A

  $(1,1)$
• B

  $(-1,-1)$
• Correct answer
  C

  $(2,2)$
• D

  $(-2,-2)$

Solution

Put $\quad x=r \cos \theta$ and $y=r \sin \theta$ $\therefore \quad r^2=x^2+y^2$ From Eqs. (i) $$ \begin{array}{lrl} & r^2-4(r \cos \theta+r \sin \theta)-4 & =0 \\ \Rightarrow \quad & x^2+y^2-4(x+y)-4 & =0 \\ \Rightarrow & x^2+y^2-4 x-4 y-4 & =0 \\ & \therefore \text { Centre of circle }(2,2) . \end{array} $$

Hence, the correct answer is option: 'C'.