If the circle x^2+y^2+2gx+2fy+c=0 (c0) touch | vedclass

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(A) The given equation of the circle is $x^2+y^2+2gx+2fy+c=0$ $(c>0)$.The center of the circle is $(-g, -f)$ and the radius is $r = \sqrt{g^2+f^2-c}$.

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$\sqrt{2c}$

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(A) The given equation of the circle is $x^2+y^2+2gx+2fy+c=0$ $(c>0)$.The center of the circle is $(-g, -f)$ and the radius is $r = \sqrt{g^2+f^2-c}$.Since the circle touches both coordinate axes,the distance from the center to the axes is equal to the radius,so $|-g| = |-f| = r = \sqrt{c}$.Since the circle lies in the third quadrant,the center must be $(-\sqrt{c}, -\sqrt{c})$.Thus,the equation of the circle is $(x+\sqrt{c})^2 + (y+\sqrt{c})^2 = c$.The circle touches the $x$-axis at $(-\sqrt{c}, 0)$ and the $y$-axis at $(0, -\sqrt{c})$.Let these points be $A(-\sqrt{c}, 0)$ and $B(0, -\sqrt{c})$.We check if these points lie on the line $x+y+\sqrt{c}=0$:For point $A$: $-\sqrt{c} + 0 + \sqrt{c} = 0$ (True).For point $B$: $0 - \sqrt{c} + \sqrt{c} = 0$ (True).Thus,the chord intercepted by the circle on the line is the segment $AB$.The length of the chord $AB$ is $\sqrt{(0 - (-\sqrt{c}))^2 + (-\sqrt{c} - 0)^2} = \sqrt{(\sqrt{c})^2 + (-\sqrt{c})^2} = \sqrt{c+c} = \sqrt{2c}$.

If the circle $x^2+y^2+2gx+2fy+c=0$ $(c>0)$ touches both the coordinate axes and lies in the third quadrant,then the length of the chord intercepted by the circle on the line $x+y+\sqrt{c}=0$ is

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