Circles
Equation of Circle
Grade None
Question:
<p>Given \ \(4a^2 + 4a + 1 = 3b^2 + 3a^2\), which of the following statements are correct about the circle with centre \ \((2, 0)\) \ and radius \ \(\sqrt{3}\)?</p><p>The equation of the circle is \ \((x-2)^2 + y^2 = 3\), i.e., \ \(x^2 + y^2 - 4x + 1 = 0\). Which of the following are true?</p>
<p>Centre of the circle is \ \((2, 0)\)</p>
<p>Radius of the circle is \ \(\sqrt{3}\)</p>
<p>The equation satisfies \ \((2a + 1)^2 = \sqrt{3}^2(a^2 + b^2)\)</p>
<p>All of the above</p>
Step-by-Step Solution
Key Concept: Recognize that 4a² + 4a + 1 = (2a+1)² and 3b² + 3a² = 3(a² + b²), then use the constraint to identify which geometric properties of the circle x² + y² - 4x + 1 = 0 hold true.
<p><strong>Step 1:</strong> Rewrite the constraint equation. Note that 4a² + 4a + 1 = (2a + 1)² and 3b² + 3a² = 3(a² + b²), so: (2a + 1)² = 3(a² + b²)</p><p><strong>Step 2:</strong> Expand: 4a² + 4a + 1 = 3a² + 3b² → a² + 4a + 1 = 3b² → a² + 4a + 1 = 3b²</p><p><strong>Step 3:</strong> Verify the given circle equation. With centre (2, 0) and radius √3: (x - 2)² + y² = 3 expands to x² - 4x + 4 + y² = 3, giving x² + y² - 4x + 1 = 0 ✓</p><p><strong>Step 4:</strong> The constraint a² + 4a + 1 = 3b² can be rewritten as: (a - 2)² + b² = 3, which is exactly the circle equation with a and b as variables. This means points (a, b) lie on the circle x² + y² - 4x + 1 = 0.</p><p><strong>Step 5:</strong> All statements about the circle's properties (equation form, centre, radius) are correct since they directly follow from the given circle definition.</p><p>∴ Answer: D</p>
Correct Answer: D