Question:

<p><strong>Step 1:</strong> Rewrite circles in standard form by completing the square.</p><p>Circle 1: $x^2 + y^2 + 2ax + 2ny = 0$ → $(x+a)^2 + (y+n)^2 = a^2 + n^2$</p><p>Center $C_1 = (-a, -n)$, Radius $r_1 = \sqrt{a^2 + n^2}$</p><p><strong>Step 2:</strong> Similarly for Circle 2: $(x+b)^2 + (y+n)^2 = b^2 + n^2$</p><p>Center $C_2 = (-b, -n)$, Radius $r_2 = \sqrt{b^2 + n^2}$</p><p><strong>Step 3:</strong> Find distance between centers.</p><p>$d = \sqrt{(-a-(-b))^2 + (-n-(-n))^2} = \sqrt{(b-a)^2} = |b-a|$</p><p><strong>Step 4:</strong> Apply tangency condition: $d = r_1 + r_2$ (external) or $d = |r_1 - r_2|$ (internal)</p><p>External: $|b-a| = \sqrt{a^2+n^2} + \sqrt{b^2+n^2}$</p><p>Internal: $|b-a| = |\sqrt{a^2+n^2} - \sqrt{b^2+n^2}|$</p><p><strong>Step 5:</strong> Square the internal tangency condition (more restrictive):</p><p>$(b-a)^2 = (\sqrt{a^2+n^2} - \sqrt{b^2+n^2})^2$</p><p>$b^2 - 2ab + a^2 = a^2 + n^2 + b^2 + n^2 - 2\sqrt{(a^2+n^2)(b^2+n^2)}$</p><p>$-2ab = 2n^2 - 2\sqrt{(a^2+n^2)(b^2+n^2)}$</p><p>$\sqrt{(a^2+n^2)(b^2+n^2)} = n^2 + ab$</p><p>$∴ a^2b^2 + a^2n^2 + b^2n^2 + n^4 = n^4 + 2abn^2 + a^2b^2$</p><p>$∴ (a-b)n = 0$ or $\boxed{a = b}$ (if circles are distinct)</p><p>For general tangency: <strong>Answer depends on options given (typically $a = b$ for internal or other relation for external)</strong></p>