Let C be a circle having centre in the first quadrant and touching the $$x$$-axis at a distance of 3 units from the origin. If the circle C has an intercept of length $$6\sqrt{3}$$ on $$y$$-axis, then the length of the chord of the circle C on the line $$x - y = 3$$ is :

If a circle touches the x-axis at $$(h, 0)$$, its center is $$(h, r)$$ and its radius is $$r$$. The length of the y-intercept cut by the circle is given by $$2\sqrt{r^2 - h^2}$$.

The length of a chord cut by a circle of radius $$r$$ on any line is $$2\sqrt{r^2 - d^2}$$, where $$d$$ is the perpendicular distance from the center to the line.

The circle touches the x-axis in the first quadrant at a distance of 3 units from the origin. Thus, the x-coordinate of the center is $$h = 3$$. Let the center of the circle be $$(3, r)$$ and its radius be $$r$$.

$$2\sqrt{r^2 - h^2} = 6\sqrt{3}$$

$$\sqrt{r^2 - 3^2} = 3\sqrt{3}$$

$$r^2 - 9 = 27 \implies r^2 = 36 \implies r = 6$$

So, the center of the circle is $$(3, 6)$$ and the radius is $$r = 6$$

$$d = \frac{|3 - 6 - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|-6|}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$$

$$\text{Length} = 2\sqrt{6^2 - (3\sqrt{2})^2} = 2\sqrt{36 - 18} = 2\sqrt{18}$$

$$\text{Length} = 2 \times 3\sqrt{2} = 6\sqrt{2}$$