Question:

Step 1: Determine the center and radius of the circle. The circle has its center on the $x$-axis at $(\gamma, 0)$ and touches the line $x + y = 0$ at the point $(2, -2)$. The radius of the circle is perpendicular to the tangent line at the point of tangency. Step 2: Calculate the coordinate $\gamma$. The slope of the tangent line $x + y = 0$ is $-1$. Since the radius is perpendicular to the tangent at the point of tangency, the slope of the radius connecting the center $(\gamma, 0)$ to the point of tangency $(2, -2)$ must be $1$. The slope of the line segment connecting $(\gamma, 0)$ and $(2, -2)$ is given by: $$m = \frac{-2 - 0}{2 - \gamma} = \frac{-2}{2 - \gamma}$$ Setting this slope equal to $1$: $$\frac{-2}{2 - \gamma} = 1$$ $$-2 = 2 - \gamma$$ $$\gamma = 4$$ Thus, the center of the circle is $(4, 0)$. Step 3: Calculate the radius of the circle. The radius $r$ is the distance between the center $(4, 0)$ and the point of tangency $(2, -2)$. $$r = \sqrt{(4 - 2)^2 + (0 - (-2))^2}$$ $$r = \sqrt{(2)^2 + (2)^2}$$ $$r = \sqrt{4 + 4}$$ $$r = \sqrt{8}$$ Step 4: Determine the greatest integral value of $\alpha$. The equation of the circle is $(x - 4)^2 + y^2 = 8$. For a point $(\alpha, \beta)$ on the circle, we have $(\alpha - 4)^2 + \beta^2 = 8$. To find the maximum value of $\alpha$, we consider the case where $\beta = 0$. $$(\alpha - 4)^2 + 0^2 = 8$$ $$(\alpha - 4)^2 = 8$$ $$\alpha - 4 = \pm\sqrt{8}$$ $$\alpha - 4 = \pm 2\sqrt{2}$$ $$\alpha = 4 \pm 2\sqrt{2}$$ The maximum value of $\alpha$ is $4 + 2\sqrt{2}$. Since $\sqrt{2} \approx 1.414$, we have $2\sqrt{2} \approx 2.828$. So, $\alpha_{\text{max}} \approx 4 + 2.828 = 6.828$. The greatest integral value of $\alpha$ is $6$. Step 5: Calculate the length of the tangent drawn from the origin to the circle. The length of the tangent $L$ from an external point $(x_1, y_1)$ to a circle $(x - h)^2 + (y - k)^2 = r^2$ is given by $L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2}$. Here, the external point is the origin $(0, 0)$, the center of the circle is $(4, 0)$, and the radius squared is $r^2 = 8$. $$L = \sqrt{(0 - 4)^2 + (0 - 0)^2 - 8}$$ $$L = \sqrt{(-4)^2 + 0^2 - 8}$$ $$L = \sqrt{16 - 8}$$ $$L = \sqrt{8}$$