Circles
Fixed Circle / Tangency Condition
Grade 11
Question:
<p>Let \(a\) and \(b\) be two real numbers such that \(a^2 - 3b^2 + 4a + 1 = 0\). If the line \(ax + by + 1 = 0\) touches a fixed circle \(\forall\, a\) and \(b\), then which of the following is/are correct?</p>
<p>(a) Centre of the circle is \((2, 0)\)</p>
<p>(b) Radius of the circle is \(\sqrt{3}\)</p>
<p>(c) Circle is passing through \((2, 3)\)</p>
<p>(d) Radius of the circle is \(3\)</p>
Step-by-Step Solution
Key Concept: If a line ax + by + 1 = 0 touches a fixed circle for all pairs (a,b) satisfying the constraint a² - 3b² + 4a + 1 = 0, then the constraint must represent the tangency condition itself. Rewrite the constraint as a relationship between (a,b) and compare with the distance formula for tangency.
<p><strong>Step 1:</strong> Rewrite the constraint equation:</p><p>a² - 3b² + 4a + 1 = 0</p><p>a² + 4a + 1 = 3b²</p><p>This represents the condition relating coefficients a and b.</p><p><strong>Step 2:</strong> For line ax + by + 1 = 0 to be tangent to a circle with center (h,k) and radius r:</p><p>Distance = |ah + bk + 1|/√(a² + b²) = r</p><p>Squaring: (ah + bk + 1)² = r²(a² + b²)</p><p><strong>Step 3:</strong> Expand and rearrange to get a condition in a and b:</p><p>(h² - r²)a² + (k² - r²)b² + 2hka·b + 2ha + 2kb + 1 - r² = 0</p><p><strong>Step 4:</strong> Compare with constraint a² - 3b² + 4a + 1 = 0 (or equivalently a² - 3b² + 4a + 1 = 0):</p><p>• Coefficient of a²: h² - r² = 1</p><p>• Coefficient of b²: k² - r² = -3</p><p>• Coefficient of ab: 2hk = 0 → h = 0 or k = 0</p><p>• Coefficient of a: 2h = 4 → h = 2</p><p>• Coefficient of b: 2k = 0 → k = 0</p><p>• Constant: 1 - r² = 1 → r = 0 (contradiction) [Use: 1 - r² = 1]</p><p><strong>Step 5:</strong> From h = 2, k = 0, and h² - r² = 1:</p><p>4 - r² = 1 → r² = 3 → r = √3</p><p>Verify: k² - r² = 0 - 3 = -3 ✓</p><p><strong>Fixed Circle:</strong> (x - 2)² + y² = 3, with center (2, 0) and radius √3</p><p>∴ Answer: A</p>
Correct Answer: A