A circle lies in the second quadrant and touc | vedclass

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(B) Since the circle lies in the second quadrant and touches both axes,its centre must be $(-r, r)$,where $r$ is the radius.Given $r = 4$,the centre i

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$x^2 + y^2 + 8x - 8y + 16 = 0$

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(B) Since the circle lies in the second quadrant and touches both axes,its centre must be $(-r, r)$,where $r$ is the radius.Given $r = 4$,the centre is $(-4, 4)$.The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - (-4))^2 + (y - 4)^2 = 4^2$.$(x + 4)^2 + (y - 4)^2 = 16$.Expanding this,we get $(x^2 + 8x + 16) + (y^2 - 8y + 16) = 16$.$x^2 + y^2 + 8x - 8y + 16 = 0$.

$A$ circle lies in the second quadrant and touches both the axes. If the radius of the circle is $4$,then its equation is

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