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(D) The given equation of the line pair is $x^2 - 2x + 1 - y^2 = 0$,which can be written as $(x - 1)^2 - y^2 = 0$.This factors into $(x - y - 1)(x + y

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$(A)$ or $(C)$ both

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(D) The given equation of the line pair is $x^2 - 2x + 1 - y^2 = 0$,which can be written as $(x - 1)^2 - y^2 = 0$.This factors into $(x - y - 1)(x + y - 1) = 0$.Thus,the two lines are $L_1: x - y - 1 = 0$ and $L_2: x + y - 1 = 0$.Let the centre of the circle be $(h, k)$. Since the circle touches these lines,the perpendicular distance from the centre to both lines must be equal to the radius $r$.$r = \frac{|h - k - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|h + k - 1|}{\sqrt{1^2 + 1^2}}$.This implies $|h - k - 1| = |h + k - 1|$.Case $1$: $h - k - 1 = h + k - 1 \implies k = 0$.Case $2$: $h - k - 1 = -(h + k - 1) \implies h - k - 1 = -h - k + 1 \implies 2h = 2 \implies h = 1$. However,if $h=1$,the lines intersect at $(1, 0)$,which is the centre,making the radius $0$,which is not a circle.So,$k = 0$ and the centre is $(h, 0)$.The radius $r = \frac{|h - 0 - 1|}{\sqrt{2}} = \frac{|h - 1|}{\sqrt{2}}$.The equation of the circle is $(x - h)^2 + y^2 = r^2 = \frac{(h - 1)^2}{2}$.The circle passes through $\left( 3, \sqrt{\frac{7}{2}} \right)$,so $(3 - h)^2 + \frac{7}{2} = \frac{(h - 1)^2}{2}$.$2(9 - 6h + h^2) + 7 = h^2 - 2h + 1$.$18 - 12h + 2h^2 + 7 = h^2 - 2h + 1$.$h^2 - 10h + 24 = 0$.$(h - 6)(h - 4) = 0$.Thus,$h = 4$ or $h = 6$.The possible centres are $(4, 0)$ and $(6, 0)$.

$A$ circle passes through the point $\left( 3, \sqrt{\frac{7}{2}} \right)$ and touches the line pair $x^2 - y^2 - 2x + 1 = 0$. The coordinates of the centre of the circle are:

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