A) \[7/2\] units
B) \[3\] units
C) \[4\] units
D) \[\sqrt{12}\] units
Correct Answer: C
Solution :
Given equation of ellipse is \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1,\] Here, \[{{a}^{2}}=16,{{b}^{2}}=9\] \[\therefore \] \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{9}{16}}\] \[=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}\] and foci \[=(\pm \,\,ae,\,\,0)\] \[=\left( \pm 4\times \frac{\sqrt{7}}{4},0 \right)=(\pm \sqrt{7},0)\] Since, circle with centre \[(0,\,\,3)\] is passing through \[(\pm \,\sqrt{7},0)\]. \[\therefore \] Required radius = distance between \[(0,\,\,3)\] and \[(\pm \,\sqrt{7},0)\] \[=\sqrt{{{(0\overset{-}{\mathop{+}}\,\sqrt{7})}^{2}}+{{(3-0)}^{2}}}\] \[=\sqrt{7+9}=\sqrt{16}=4units\]
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