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

Question

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

Aicharya · Accepted Answer

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

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

Solution