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If the circle $x^2+y^2+2 g x+2 f y+c=0(c>0)$ 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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Verified Answer
The correct answer is:
$\sqrt{2 \mathrm{C}}$
Given equation of circle
$$
x^2+y^2+2 g x+2 f y+c=0 \quad(c>0)
$$
Coordinate of centre $=(-g,-f)$
$$
\text { radius }=\sqrt{g^2+f^2-c}
$$
Circle touch both the axes, so
$$
\begin{aligned}
g^2 & =f^2=c \Rightarrow g= \pm \sqrt{c} \\
f & = \pm \sqrt{c}
\end{aligned}
$$
So, coordinate of centre lies in third quadrant so, centre is $(-\sqrt{c},-\sqrt{c})$
equation of given line $x+y+\sqrt{c}=0$
So, line pass through the point of contact to axes of the circle.
$$
\text { Hence, intercept } \begin{aligned}
A B & =\sqrt{(-\sqrt{c}-0)^2+(0+\sqrt{c})^2} \\
& =\sqrt{c+c}=\sqrt{2 c}
\end{aligned}
$$
$$
x^2+y^2+2 g x+2 f y+c=0 \quad(c>0)
$$
Coordinate of centre $=(-g,-f)$
$$
\text { radius }=\sqrt{g^2+f^2-c}
$$
Circle touch both the axes, so
$$
\begin{aligned}
g^2 & =f^2=c \Rightarrow g= \pm \sqrt{c} \\
f & = \pm \sqrt{c}
\end{aligned}
$$
So, coordinate of centre lies in third quadrant so, centre is $(-\sqrt{c},-\sqrt{c})$
equation of given line $x+y+\sqrt{c}=0$
So, line pass through the point of contact to axes of the circle.
$$
\text { Hence, intercept } \begin{aligned}
A B & =\sqrt{(-\sqrt{c}-0)^2+(0+\sqrt{c})^2} \\
& =\sqrt{c+c}=\sqrt{2 c}
\end{aligned}
$$
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