Key Concepts and Formulas

• Equation of a circle: The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$. The general equation is $x^2 + y^2 + 2gx + 2fy + c = 0$, with center $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$.
• Conditions for circles to touch: Two circles with centers $C_1, C_2$ and radii $r_1, r_2$ touch each other if the distance between their centers $C_1C_2$ is equal to the sum of their radii (external touching) or the absolute difference of their radii (internal touching).
  • External touching: $C_1C_2 = r_1 + r_2$
  • Internal touching: $C_1C_2 = |r_1 - r_2|$
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Step 1: Find the center and radius of the first circle

The equation of the first circle is $x^2 + y^2 = ax$. We rewrite it in the standard form to find its center and radius. $x^2 - ax + y^2 = 0$ Completing the square for the $x$ terms, we add and subtract $(a/2)^2$: $x^2 - ax + \left(\frac{a}{2}\right)^2 + y^2 = \left(\frac{a}{2}\right)^2$ $\left(x - \frac{a}{2}\right)^2 + y^2 = \left(\frac{a}{2}\right)^2$ Thus, the center of the first circle $C_1$ is $\left(\frac{a}{2}, 0\right)$ and its radius $r_1$ is $\left|\frac{a}{2}\right|$.

Step 2: Find the center and radius of the second circle

The equation of the second circle is $x^2 + y^2 = c^2$. This is in the standard form $(x - 0)^2 + (y - 0)^2 = c^2$. Thus, the center of the second circle $C_2$ is $(0, 0)$ and its radius $r_2$ is $c$.

Step 3: Calculate the distance between the centers

The distance between the centers $C_1\left(\frac{a}{2}, 0\right)$ and $C_2(0, 0)$ is: $C_1C_2 = \sqrt{\left(\frac{a}{2} - 0\right)^2 + (0 - 0)^2} = \sqrt{\left(\frac{a}{2}\right)^2} = \left|\frac{a}{2}\right|$

Step 4: Apply the touching condition

Since the circles touch each other, the distance between their centers must be equal to either the sum or the absolute difference of their radii: $C_1C_2 = r_1 + r_2 \quad \text{or} \quad C_1C_2 = |r_1 - r_2|$ Substituting the values we found: $\left|\frac{a}{2}\right| = \left|\frac{a}{2}\right| + c \quad \text{or} \quad \left|\frac{a}{2}\right| = \left| \left|\frac{a}{2}\right| - c \right|$

The first case, $\left|\frac{a}{2}\right| = \left|\frac{a}{2}\right| + c$, implies $c = 0$, which contradicts the given condition $c > 0$. Therefore, we consider the second case: $\left|\frac{a}{2}\right| = \left| \left|\frac{a}{2}\right| - c \right|$ Squaring both sides: $\left(\frac{a}{2}\right)^2 = \left( \left|\frac{a}{2}\right| - c \right)^2$ $\frac{a^2}{4} = \left(\frac{|a|}{2} - c\right)^2$ Taking the square root of both sides: $\left|\frac{a}{2}\right| = \left| \left|\frac{a}{2}\right| - c \right|$ This means either $\frac{a}{2} = \frac{|a|}{2} - c$ or $\frac{a}{2} = -(\frac{|a|}{2} - c) = c - \frac{|a|}{2}$.

Case 1: $\frac{|a|}{2} - c = \frac{a}{2}$ If $a \ge 0$, then $\frac{a}{2} - c = \frac{a}{2}$, which gives $c = 0$, a contradiction. If $a < 0$, then $-\frac{a}{2} - c = \frac{a}{2}$, which gives $-a = c$, so $a = -c$. Thus, $|a| = c$.

Case 2: $c - \frac{|a|}{2} = \frac{a}{2}$ If $a \ge 0$, then $c - \frac{a}{2} = \frac{a}{2}$, which gives $c = a$, so $|a| = c$. If $a < 0$, then $c - \frac{-a}{2} = \frac{a}{2}$, which gives $c + \frac{a}{2} = \frac{a}{2}$, so $c = 0$, a contradiction.

In both valid sub-cases, we have $|a| = c$.

Common Mistakes & Tips

• Remember to consider both external and internal touching conditions.
• Don't forget the absolute value when taking the square root of a squared term.
• Be careful with signs when dealing with the distance formula and completing the square.

Summary

We found the centers and radii of the two circles. Then, we used the condition for the circles to touch, considering both external and internal touching. By carefully analyzing the resulting equations, we arrived at the condition $|a| = c$.

Final Answer

The final answer is \boxed{| a | = c}, which corresponds to option (A).