<p>A circle passing through origin O cuts two straight lines \(x - y = 0\) and \(x + y = 0\) in points A and B respectively. If abscissae of A and B are roots of the equation \(x^2 + ax + b = 0\), then the equation of the given circle is:</p>
<p>(a) \(x^2 + y^2 + ax - by = 0\)</p>
<p>(b) \(x^2 + y^2 - x\sqrt{4b - a^2} + yb = 0\)</p>
<p>(c) \(x^2 + y^2 + ax \pm y\sqrt{a^2 - 4b} = 0\)</p>
<p>(d) \(x^2 + y^2 - ax \pm y\sqrt{a^2 - 4b} = 0\)</p>
Step-by-Step Solution
Key Concept: Since the circle passes through origin O and intersects two perpendicular lines x - y = 0 and x + y = 0 at points A and B respectively, the abscissae of A and B are roots of x² + ax + b = 0. Use the circle equation through origin and apply coordinate constraints to find the relationship between circle parameters and quadratic coefficients.
<p><strong>Step 1:</strong> Set up point coordinates. Since A lies on x - y = 0, let A = (x₁, x₁). Since B lies on x + y = 0, let B = (x₂, -x₂), where x₁ and x₂ are roots of x² + ax + b = 0.</p><p><strong>Step 2:</strong> Use Vieta's formulas. From x² + ax + b = 0: x₁ + x₂ = -a and x₁x₂ = b.</p><p><strong>Step 3:</strong> Write the general circle equation passing through origin: x² + y² + 2gx + 2fy = 0 (constant term = 0 since circle passes through O).</p><p><strong>Step 4:</strong> Since A(x₁, x₁) lies on the circle: x₁² + x₁² + 2gx₁ + 2fx₁ = 0 → 2x₁² + 2(g+f)x₁ = 0 → x₁[x₁ + (g+f)] = 0. Since x₁ ≠ 0, we get x₁ = -(g+f).</p><p><strong>Step 5:</strong> Since B(x₂, -x₂) lies on the circle: x₂² + x₂² + 2gx₂ + 2f(-x₂) = 0 → 2x₂² + 2(g-f)x₂ = 0 → x₂[x₂ + (g-f)] = 0. Since x₂ ≠ 0, we get x₂ = -(g-f).</p><p><strong>Step 6:</strong> From Steps 4 and 5: x₁ = -(g+f) and x₂ = -(g-f). Adding: x₁ + x₂ = -2g → -2g = -a → g = a/2. Subtracting: x₁ - x₂ = -2f → -2f = x₁ - x₂.</p><p><strong>Step 7:</strong> Calculate x₁ - x₂. We know (x₁ - x₂)² = (x₁ + x₂)² - 4x₁x₂ = (-a)² - 4b = a² - 4b. Therefore x₁ - x₂ = ±√(a² - 4b), giving -2f = ±√(a² - 4b) → f = ∓√(a² - 4b)/2.</p><p><strong>Step 8:</strong> The circle equation is x² + y² + 2gx + 2fy = 0 → x² + y² + ax + 2f·y = 0. Substituting f = ∓√(a² - 4b)/2: x² + y² + ax ∓ y√(a² - 4b) = 0, or equivalently x² + y² - ax ± y√(a² - 4b) = 0 (absorbing the sign into ±).</p><p><strong>∴ Answer:</strong> d</p>
Correct Answer: d