Step-by-step Solution:

Problem Statement
Find the condition for the circles \( x^2 + y^2 + c^2 = 2ax \) and \( x^2 + y^2 + c^2 = 2by \) to touch each other externally.
Solution
Step 1: Rewrite in Standard Form
1. First Circle: \[ x^2 - 2ax + y^2 = -c^2 \] Complete the square: \[ (x - a)^2 + y^2 = a^2 - c^2 \] Center: \((a, 0)\)
Radius: \( r_1 = \sqrt{a^2 - c^2} \)
2. Second Circle: \[ x^2 + y^2 - 2by = -c^2 \] Complete the square: \[ x^2 + (y - b)^2 = b^2 - c^2 \] Center: \((0, b)\)
Radius: \( r_2 = \sqrt{b^2 - c^2} \)
Step 2: Condition for External Tangency
Two circles touch externally if the distance between centers equals the sum of radii: \[ \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2} \] Square both sides: \[ a^2 + b^2 = (a^2 - c^2) + (b^2 - c^2) + 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \] Simplify: \[ 2c^2 = 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \] Square again: \[ c^4 = (a^2 - c^2)(b^2 - c^2) \] Expand and simplify: \[ a^2b^2 = c^2(a^2 + b^2) \] Divide by \( a^2b^2c^2 \): \[ \frac{1}{c^2} = \frac{1}{a^2} + \frac{1}{b^2} \] Final Answer
The circles touch externally if: \[ {\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}} \]