<p>The radius of circle passing through both the foci of ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and having centre at $(0, 3)$ is $r$, then $r = $ ?</p>
Step-by-Step Solution
Key Concept: For an ellipse, find the foci using $c^2 = a^2 - b^2$, then calculate the distance from the given centre to a focus.
<p><strong>Step 1:</strong> For ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$, we have $a^2 = 16$ and $b^2 = 9$.</p><p><strong>Step 2:</strong> Calculate $c^2 = a^2 - b^2 = 16 - 9 = 7$, so $c = \sqrt{7}$.</p><p><strong>Step 3:</strong> The foci are at $(\pm\sqrt{7}, 0)$.</p><p><strong>Step 4:</strong> Circle with centre at $(0, 3)$ passing through focus $(\sqrt{7}, 0)$: $r = \sqrt{(\sqrt{7})^2 + (0-3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4$.</p><p>∴ Answer is $q$ (4).</p>
Correct Answer: q