<p>The radius of the circle passing through the foci of the ellipse \(\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1\) and having its centre at (0, 3), is</p>
Step-by-Step Solution
Key Concept: The foci of the ellipse lie on the major axis (x-axis), and the radius of a circle with center (0,3) passing through both foci is the distance from (0,3) to either focus.
<p><strong>Step 1:</strong> Identify the ellipse parameters. From $\frac{x^2}{16} + \frac{y^2}{9} = 1$, we have $a^2 = 16$ and $b^2 = 9$, so $a = 4$ and $b = 3$.</p><p><strong>Step 2:</strong> Since $a^2 > b^2$, the major axis is along the x-axis. Calculate $c$: $c^2 = a^2 - b^2 = 16 - 9 = 7$, so $c = \sqrt{7}$.</p><p><strong>Step 3:</strong> The foci are at $(\pm\sqrt{7}, 0)$.</p><p><strong>Step 4:</strong> Find the radius of the circle with center at $(0, 3)$ passing through $(\sqrt{7}, 0)$:</p><p>$$r = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4$$</p><p><strong>Step 5:</strong> Verify with the other focus $(-\sqrt{7}, 0)$: $r = \sqrt{7 + 9} = 4$ ✓</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: A