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(C) The given lines are $L_1: 3x - 4y + 1 = 0$ and $L_2: 4x + 3y - 7 = 0$. Note that these lines are perpendicular since their slopes $m_1 = 3/4$ and

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Both $(a)$ and $(b)$

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(C) The given lines are $L_1: 3x - 4y + 1 = 0$ and $L_2: 4x + 3y - 7 = 0$. Note that these lines are perpendicular since their slopes $m_1 = 3/4$ and $m_2 = -4/3$ satisfy $m_1 	imes m_2 = -1$.The angle bisectors of these lines are given by $\frac{3x - 4y + 1}{\sqrt{3^2 + (-4)^2}} = \pm \frac{4x + 3y - 7}{\sqrt{4^2 + 3^2}}$,which simplifies to $3x - 4y + 1 = \pm(4x + 3y - 7)$.Case $1$: $3x - 4y + 1 = 4x + 3y - 7 \implies x + 7y - 8 = 0$.Case $2$: $3x - 4y + 1 = -4x - 3y + 7 \implies 7x - y - 6 = 0$.The center $(h, k)$ of the circle lies on these bisectors and the distance from the center to the lines equals the radius $r$. By verifying the given options,both circles satisfy the condition of passing through $(2, 3)$ and being tangent to the given lines.

The equations of the circles which touch the lines $3x - 4y + 1 = 0$ and $4x + 3y - 7 = 0$ and pass through the point $(2, 3)$ are:

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