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(B) The equation of the circle is $x^{2} + y^{2} - 2x + 2fy + 1 = 0$. The centre of the circle is $(1, -f)$.Since the diameters pass through the centr

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$2$

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(B) The equation of the circle is $x^{2} + y^{2} - 2x + 2fy + 1 = 0$. The centre of the circle is $(1, -f)$.Since the diameters pass through the centre,we substitute $(1, -f)$ into the equations of the diameters:$2p(1) - (-f) = 1 \implies 2p + f = 1 \implies f = 1 - 2p$.$2(1) + p(-f) = 4p \implies 2 - pf = 4p$.Substituting $f = 1 - 2p$ into the second equation:$2 - p(1 - 2p) = 4p \implies 2 - p + 2p^{2} = 4p \implies 2p^{2} - 5p + 2 = 0$.Solving for $p$: $(2p - 1)(p - 2) = 0$,so $p = 1/2$ or $p = 2$.If $p = 1/2$,$f = 1 - 2(1/2) = 0$. Centre is $(1, 0)$.If $p = 2$,$f = 1 - 2(2) = -3$. Centre is $(1, 3)$.The hyperbola is $3x^{2} - y^{2} = 3$,which is $x^{2} - y^{2}/3 = 1$. The tangent line $y = mx + c$ has $c^{2} = a^{2}m^{2} - b^{2} = m^{2} - 3$.Case $1$: Centre $(1, 0)$. $0 = m(1) + c \implies c = -m$. Then $c^{2} = m^{2} = m^{2} - 3$,which is impossible.Case $2$: Centre $(1, 3)$. $3 = m(1) + c \implies c = 3 - m$. Then $c^{2} = (3 - m)^{2} = m^{2} - 3$.$9 - 6m + m^{2} = m^{2} - 3 \implies 6m = 12 \implies m = 2$.

Let the equation of two diameters of a circle $x^{2} + y^{2} - 2x + 2fy + 1 = 0$ be $2px - y = 1$ and $2x + py = 4p$. Then the slope $m \in (0, \infty)$ of the tangent to the hyperbola $3x^{2} - y^{2} = 3$ passing through the centre of the circle is equal to $......$

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