The circle x^2 + y^2 - 8x + 4y + 4 = 0 touche | vedclass

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(B) The given equation is $x^2 + y^2 - 8x + 4y + 4 = 0$.Rewriting the equation by completing the square:$(x^2 - 8x + 16) + (y^2 + 4y + 4) = -4 + 16 +

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$y$-axis only

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(B) The given equation is $x^2 + y^2 - 8x + 4y + 4 = 0$.Rewriting the equation by completing the square:$(x^2 - 8x + 16) + (y^2 + 4y + 4) = -4 + 16 + 4$$(x - 4)^2 + (y + 2)^2 = 16$$(x - 4)^2 + (y + 2)^2 = 4^2$.The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre and $r$ is the radius.Comparing the equations,we get the centre $(h, k) = (4, -2)$ and radius $r = 4$.$A$ circle touches the $x$-axis if $|k| = r$ and the $y$-axis if $|h| = r$.Here,$|h| = |4| = 4 = r$,so the circle touches the $y$-axis.Also,$|k| = |-2| = 2 
eq r$,so it does not touch the $x$-axis.Therefore,the circle touches the $y$-axis only.

The circle $x^2 + y^2 - 8x + 4y + 4 = 0$ touches

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