Let 6 and 8 be the X and Y-intercepts made by | vedclass

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(D) The $X$-intercept is $2\sqrt{g^2-c} = 6 \Rightarrow g^2-c = 9$ ...$(1)$ The $Y$-intercept is $2\sqrt{f^2-c} = 8 \Rightarrow f^2-c = 16$ ...$(2)$ T

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$5$

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(D) The $X$-intercept is $2\sqrt{g^2-c} = 6 \Rightarrow g^2-c = 9$ ...$(1)$ The $Y$-intercept is $2\sqrt{f^2-c} = 8 \Rightarrow f^2-c = 16$ ...$(2)$ The line $gx+fy+1=0$ passes through $(1, -1)$,so $g(1) + f(-1) + 1 = 0 \Rightarrow g-f = -1$ ...$(3)$ Subtracting $(2)$ from $(1)$: $g^2-f^2 = -7 \Rightarrow (g-f)(g+f) = -7$. Substituting $g-f = -1$,we get $g+f = 7$ ...$(4)$ Solving $(3)$ and $(4)$: $2g = 6 \Rightarrow g = 3$ and $f = 4$. From $(1)$,$c = g^2-9 = 3^2-9 = 0$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{3^2+4^2-0} = \sqrt{9+16} = 5$.

Let $6$ and $8$ be the $X$ and $Y$-intercepts made by the circle $S \equiv x^2+y^2+2gx+2fy+c=0$ respectively. If $gx+fy+1=0$ is a line passing through the point $(1, -1)$,then the radius of the circle $S=0$ is

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