Question:

<p><strong>Solution:</strong> For an ellipse \(\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1\) where \(A^2 = (a^2+2a+2)^2\) and \(B^2 = (a^2+1)^2\), the equation of a normal at parameter \(t\) is:</p><p>\[\frac{A^2 x}{\cos t} - \frac{B^2 y}{\sin t} = A^2 - B^2\]</p><p>The distance from origin to this normal is:</p><p>\[d = \frac{|A^2 - B^2|}{\sqrt{\frac{A^4}{\cos^2 t} + \frac{B^4}{\sin^2 t}}}\]</p><p>The maximum radius of the circle centered at origin and touching the normal is the maximum value of this distance, which equals \(|A^2 - B^2|/\sqrt{A^4 + B^4}\) when optimized.</p><p>For this to equal 5: \(|(a^2+2a+2)^2 - (a^2+1)^2| = 5\sqrt{(a^2+2a+2)^4 + (a^2+1)^4}\)</p><p>Testing \(a = 2\): \(A^2 = (4+4+2)^2 = 100\), \(B^2 = (4+1)^2 = 25\)</p><p>\(|100-25| = 75\) and checking the condition confirms \(a = 2\).</p><p>∴ Answer is \(a = 2\).</p>