Circle(s) touching $x$-axis at a distance $3$ from the origin and having an intercept of length $2\sqrt{7}$ on $y$-axis is (are) (1) $x^{2} + y^{2} - 6x + 8y + 9 = 0$ (2) $x^{2} + y^{2} - 6x + 7y + 9 = 0$ (3) $x^{2} + y^{2} - 6x - 8y + 9 = 0$ (4) $x^{2} + y^{2} - 6x - 7y + 9 = 0$

$\text{Given: circle touches the } x\text{-axis at distance } 3 \text{ from origin}$ $\text{intercept on } y\text{-axis} = 2\sqrt{7}$ $\text{To find: which of the given circles satisfy both conditions}$ $1.\ \text{fix the point of contact and the centre}$ $2.\ \text{write the general equation}$ $3.\ \text{use the } y\text{-axis intercept, then match options}$ $\text{point of contact } = (3,\, 0)$ $\text{centre} = (3,\, \pm r)$ $(x-3)^2 + (y \pm r)^2 = r^2$ $x^2 + y^2 - 6x + c\,y + 9 = 0, \quad |c| = 2r$ $x = 0:\quad y^2 + c\,y + 9 = 0$ $\text{the intercepts are the roots } y_1, y_2$ $|y_1 - y_2| = \sqrt{(y_1+y_2)^2 - 4y_1 y_2}$ $y_1 + y_2 = -c, \quad y_1 y_2 = 9$ $|y_1 - y_2| = \sqrt{c^2 - 36}$ $\sqrt{c^2 - 36} = 2\sqrt{7}$ $c^2 - 36 = 28 \ \Rightarrow \ c^2 = 64 \ \Rightarrow \ c = \pm 8$ $c = \pm 8 \ \Rightarrow \ \text{options (1) and (3)}$ $c = \pm 7 \ \Rightarrow \ \sqrt{49-36} = \sqrt{13} \neq 2\sqrt{7} \ \Rightarrow \ \text{reject (2), (4)}$ $x^2 + y^2 - 6x + 8y + 9 = 0$ $\text{centre } (3,-4), \quad r = \sqrt{9+16-9} = 4$ $x^2 + y^2 - 6x - 8y + 9 = 0$ $\text{centre } (3,4), \quad r = 4$ $\text{Both: } r = 4, \text{ touch } x\text{-axis at } (3,0)$ $\text{and cut } y\text{-intercept } 2\sqrt{7}$ $\boxed{\text{Answer: (1) and (3)}}$