The centre of the circle r^2-4r(cos theta+sin | vedclass

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(C) Given the equation of the circle in polar form: $r^2-4r(\cos 	heta+\sin 	heta)-4=0$ $(i)$We know that $x=r \cos 	heta$,$y=r \sin 	heta$,and $r

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$(2,2)$

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(C) Given the equation of the circle in polar form: $r^2-4r(\cos 	heta+\sin 	heta)-4=0$ $(i)$We know that $x=r \cos 	heta$,$y=r \sin 	heta$,and $r^2=x^2+y^2$.Substituting these into equation $(i)$:$x^2+y^2-4(x+y)-4=0$$x^2+y^2-4x-4y-4=0$Comparing this with the general equation of a circle $x^2+y^2+2gx+2fy+c=0$,we get:$2g = -4 \Rightarrow g = -2$$2f = -4 \Rightarrow f = -2$The centre of the circle is $(-g, -f) = (-(-2), -(-2)) = (2, 2)$.

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

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