The radius of the circle passing through the foci of the ellipse <span class="math-tex">\(\dfrac{x^2}{16}+ \dfrac{y^2}{9}=1\)</span> and having it centre at (0, 3) is
For the ellipse, \(a^2 = 16\), \(b^2 = 9\), so \(c = \sqrt{16 - 9} = \sqrt{7}\). The foci are \((\pm\sqrt{7}, 0)\).
Radius = distance from (0, 3) to \((\sqrt{7}, 0)\) = \(\sqrt{7 + 9} = 4\).
Answer: (A) 4 units
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