If the circle x^{2}+y^{2}-2gx+6y-19c=0,where | vedclass

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(D) The equation of the circle is $x^{2}+y^{2}-2gx+6y-19c=0$. Since the circle passes through $(6,1)$,we have: $6^{2}+1^{2}-2g(6)+6(1)-19c=0$ $36+1-12

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$2\sqrt{23}$

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(D) The equation of the circle is $x^{2}+y^{2}-2gx+6y-19c=0$. Since the circle passes through $(6,1)$,we have: $6^{2}+1^{2}-2g(6)+6(1)-19c=0$ $36+1-12g+6-19c=0$ $43-12g-19c=0 \implies 12g+19c=43$ $(1)$ The centre of the circle is $(g, -3)$. Since the centre lies on the line $x-2cy=8$,we have: $g-2c(-3)=8 \implies g+6c=8$ $(2)$ From $(2)$,$g=8-6c$. Substituting into $(1)$: $12(8-6c)+19c=43$ $96-72c+19c=43$ $-53c = -53 \implies c=1$ Substituting $c=1$ into $g=8-6c$,we get $g=8-6=2$. The equation of the circle is $x^{2}+y^{2}-4x+6y-19=0$. The length of the intercept on the $x$-axis is given by $2\sqrt{g^{2}-c'}$ where $c'$ is the constant term $-19$. Length $= 2\sqrt{g^{2}-(-19)} = 2\sqrt{2^{2}+19} = 2\sqrt{4+19} = 2\sqrt{23}$.

If the circle $x^{2}+y^{2}-2gx+6y-19c=0$,where $g, c \in R$,passes through the point $(6,1)$ and its centre lies on the line $x-2cy=8$,then the length of the intercept made by the circle on the $x$-axis is:

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