<p>Foci of ellipse \(\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1\) are given by \((\pm ae, 0)\). The radius of the circle through a focus and having centre at \(C(0, 3)\) is:</p>
Step-by-Step Solution
Key Concept: For the ellipse x²/16 + y²/9 = 1, identify a² = 16 and b² = 9, then calculate eccentricity e = √(1 - b²/a²) to find the foci at (±ae, 0). The radius is the distance from center C(0,3) to a focus point.
<p><strong>Step 1:</strong> From the ellipse equation x²/16 + y²/9 = 1, we identify a² = 16 and b² = 9, so a = 4 and b = 3.</p><p><strong>Step 2:</strong> Calculate eccentricity: e² = 1 - b²/a² = 1 - 9/16 = 7/16, so e = √7/4.</p><p><strong>Step 3:</strong> The foci are at (±ae, 0) = (±4·√7/4, 0) = (±√7, 0).</p><p><strong>Step 4:</strong> The radius of the circle with center C(0, 3) passing through focus F(√7, 0) is:</p><p>r = √[(√7 - 0)² + (0 - 3)²] = √[7 + 9] = √16 = 4</p><p>∴ Answer: A (radius = 4)</p>
Correct Answer: A