The circle 4x^2+4y^2-12x-12y+9=0 | vedclass

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(A) The given equation of the circle is $4x^2+4y^2-12x-12y+9=0$. Dividing by $4$,we get $x^2+y^2-3x-3y+\frac{9}{4}=0$. Rearranging the terms,we have $

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touches both the axes

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(A) The given equation of the circle is $4x^2+4y^2-12x-12y+9=0$. Dividing by $4$,we get $x^2+y^2-3x-3y+\frac{9}{4}=0$. Rearranging the terms,we have $(x^2-3x)+(y^2-3y)=-\frac{9}{4}$. Completing the square,we get $(x-\frac{3}{2})^2-\frac{9}{4}+(y-\frac{3}{2})^2-\frac{9}{4}=-\frac{9}{4}$. This simplifies to $(x-\frac{3}{2})^2+(y-\frac{3}{2})^2=\frac{9}{4}$,which is $(x-\frac{3}{2})^2+(y-\frac{3}{2})^2=(\frac{3}{2})^2$. Comparing this with the standard form $(x-h)^2+(y-k)^2=r^2$,the centre is $(\frac{3}{2}, \frac{3}{2})$ and the radius is $r=\frac{3}{2}$. Since the distance of the centre from both axes is equal to the radius (i.e.,$|h|=|k|=r=\frac{3}{2}$),the circle touches both the axes.

The circle $4x^2+4y^2-12x-12y+9=0$

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