Key Concepts and Formulas

• Equation of a Circle: The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$, with center $(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$.
• Circle Not Intersecting Axes: A circle neither intersects nor touches the coordinate axes if its radius is less than the absolute values of both coordinates of its center: $r < |h|$ and $r < |k|$, where $(h, k)$ is the center.
• Point Inside a Circle: A point $(x_1, y_1)$ lies inside the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ if $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c < 0$.

Step-by-Step Solution

Step 1: Convert the Circle Equation to Standard Form

We start with the equation $36x^2 + 36y^2 - 108x + 120y + C = 0$. To obtain the standard form, divide by 36: $x^2 + y^2 - 3x + \frac{10}{3}y + \frac{C}{36} = 0$ Thus, $2g = -3$, $2f = \frac{10}{3}$, and $c = \frac{C}{36}$.

Step 2: Determine the Center and Radius of the Circle

The center of the circle is $(-g, -f) = \left(\frac{3}{2}, -\frac{5}{3}\right)$. The radius is $r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(\frac{5}{3}\right)^2 - \frac{C}{36}} = \sqrt{\frac{9}{4} + \frac{25}{9} - \frac{C}{36}} = \sqrt{\frac{81 + 100 - C}{36}} = \sqrt{\frac{181 - C}{36}}$.

Step 3: Apply the Condition for the Circle Not Intersecting the Axes

Since the circle neither intersects nor touches the axes, the radius must be less than the absolute values of both coordinates of the center: $r < \left|\frac{3}{2}\right| \text{ and } r < \left|-\frac{5}{3}\right|$ $r < \frac{3}{2} \text{ and } r < \frac{5}{3}$ Since $\frac{3}{2} < \frac{5}{3}$, we use the stricter condition $r < \frac{3}{2}$. Substituting the expression for $r$, we have $\sqrt{\frac{181 - C}{36}} < \frac{3}{2}$ Squaring both sides gives $\frac{181 - C}{36} < \frac{9}{4}$ $181 - C < \frac{9}{4} \cdot 36 = 9 \cdot 9 = 81$ $181 - C < 81$ $C > 181 - 81 = 100$

Step 4: Find the Intersection Point of the Lines

We have the lines $x - 2y = 4$ and $2x - y = 5$. Multiply the first equation by 2 to get $2x - 4y = 8$. Subtracting the second equation from this gives $(2x - 4y) - (2x - y) = 8 - 5$ $-3y = 3$ $y = -1$ Substituting into $x - 2y = 4$ gives $x - 2(-1) = 4$, so $x + 2 = 4$, which means $x = 2$. The intersection point is $(2, -1)$.

Step 5: Apply the Condition for the Point Lying Inside the Circle

The point $(2, -1)$ lies inside the circle, so $2^2 + (-1)^2 - 3(2) + \frac{10}{3}(-1) + \frac{C}{36} < 0$ $4 + 1 - 6 - \frac{10}{3} + \frac{C}{36} < 0$ $-1 - \frac{10}{3} + \frac{C}{36} < 0$ $-\frac{3}{3} - \frac{10}{3} + \frac{C}{36} < 0$ $-\frac{13}{3} + \frac{C}{36} < 0$ $\frac{C}{36} < \frac{13}{3}$ $C < \frac{13}{3} \cdot 36 = 13 \cdot 12 = 156$

Step 6: Combine the Inequalities to Find the Range of C

We have $C > 100$ and $C < 156$, so $100 < C < 156$. However, we need to match the given correct answer. Let's analyze the options. Option A says $\frac{25}{9} < C < \frac{13}{3}$, which is approximately $2.78 < C < 4.33$. This clearly contradicts $C > 100$.

It appears there's an error somewhere. The range for $C$ we derived, $100 < C < 156$, does not lead to option (A). Let's work backwards from the correct answer $\frac{25}{9} < C < \frac{13}{3}$. This implies $C$ is a small number. This is where the error lies. Our derivation is correct, but the given "Correct Answer" is wrong. We will proceed with our solution.

Common Mistakes & Tips

• Remember to divide the entire circle equation by the coefficient of $x^2$ and $y^2$ to get the standard form.
• When squaring inequalities, make sure both sides are positive to avoid changing the inequality sign.
• Pay close attention to the wording of the problem, especially phrases like "neither intersects nor touches," to apply the correct conditions.

Summary

We converted the circle equation to standard form, found its center and radius, and applied the conditions for the circle not intersecting the coordinate axes and the point lying inside the circle. Combining the resulting inequalities, we found the range for $C$ to be $100 < C < 156$, which corresponds to option (D). However, the problem statement claims option (A) is correct. Our derivation is sound, suggesting the provided "Correct Answer" is incorrect.

Final Answer

The final answer is \boxed{100 < C < 156}, which corresponds to option (D).