<p>Given the equation of a circle \(x^2 + y^2 - 6x = 0\) and a parabola \(y^2 = 4x\). Find the equations of the common tangents to the circle and the parabola.</p>
<p>\(\sqrt{3}y = x + 3\) and \(\sqrt{3}y = -(x+3)\)</p>
<p>\(\sqrt{3}y = x - 3\) and \(\sqrt{3}y = -(x-3)\)</p>
<p>\(y = x + 3\) and \(y = -(x+3)\)</p>
<p>\(y = \sqrt{3}x + 1\) and \(y = -\sqrt{3}x - 1\)</p>
Step-by-Step Solution
Key Concept: A common tangent must satisfy the tangency condition for both curves simultaneously. For the parabola y²=4x, any tangent has form y=mx+1/m; this tangent must also touch the circle (x-3)²+y²=9 with distance from center equal to radius.
<p><strong>Step 1:</strong> Rewrite the circle: x² + y² - 6x = 0 → (x-3)² + y² = 9. Center C(3,0), radius r = 3.</p><p><strong>Step 2:</strong> For parabola y² = 4x (where a=1), any tangent (non-vertical) has form: <strong>y = mx + 1/m</strong>.</p><p><strong>Step 3:</strong> For this line to be tangent to circle, distance from C(3,0) to line mx - y + 1/m = 0 must equal 3:</p><p>$$\frac{|3m - 0 + 1/m|}{\sqrt{m^2 + 1}} = 3$$</p><p><strong>Step 4:</strong> Simplify: |3m + 1/m| = 3√(m² + 1)</p><p>Squaring: (3m + 1/m)² = 9(m² + 1)</p><p>9m² + 6 + 1/m² = 9m² + 9</p><p>1/m² = 3 → m² = 1/3 → m = ±1/√3</p><p><strong>Step 5:</strong> For m = 1/√3: y = x/√3 + √3 → <strong>x - √3y + 3 = 0</strong></p><p>For m = -1/√3: y = -x/√3 - √3 → <strong>x + √3y + 3 = 0</strong></p><p><strong>Step 6:</strong> Also check vertical tangent x = 0 (touches parabola at origin, distance from C is 3 ✓)</p><p>∴ <strong>Common tangents: x = 0, x - √3y + 3 = 0, and x + √3y + 3 = 0</strong></p>
Correct Answer: A