(C) Given the polar equation of the circle is $r^2-4r(\cos \theta+\sin \theta)-4=0$. We know that $x=r \cos \theta$,$y=r \sin \theta$,and $r^2=x^2+y^2

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

Class 11 Mathematics 10-1.Circle and System of Circles Equations of circle

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

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A
$(1,1)$
B
$(-1,-1)$
C
$(2,2)$
D
$(-2,-2)$

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10-1.Circle and System of Circles

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Equations of circle

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