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 the entire equation by $4$,we get:$x^2 + y^2 - 3x - 3y + \frac{9}{4

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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 the entire equation by $4$,we get:$x^2 + y^2 - 3x - 3y + \frac{9}{4} = 0$Rearranging the terms to complete the square:$(x^2 - 3x) + (y^2 - 3y) = -\frac{9}{4}$Adding $(\frac{3}{2})^2$ to both sides for both $x$ and $y$ terms:$(x^2 - 3x + \frac{9}{4}) + (y^2 - 3y + \frac{9}{4}) = -\frac{9}{4} + \frac{9}{4} + \frac{9}{4}$$(x - \frac{3}{2})^2 + (y - \frac{3}{2})^2 = (\frac{3}{2})^2$Comparing this with the standard form of a circle $(x - h)^2 + (y - k)^2 = r^2$,we have the centre $(h, k) = (\frac{3}{2}, \frac{3}{2})$ and the radius $r = \frac{3}{2}$.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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