question_answer

The radius of the largest circle, having centre \[(1,0),\] that can be inscribed in the ellipse \[{{x}^{2}}+4{{y}^{2}}=16,\]is

A)  \[\sqrt{11}\]        

B)  \[\sqrt{\frac{11}{3}}\]

C)  \[\sqrt{\frac{22}{3}}\]

D)  \[\sqrt{22}\]

Correct Answer: B

Solution :

Equation of ellipse is \[{{x}^{2}}+4{{y}^{2}}=16\] Let equation of circle whose centre \[(1,\,0)\] and r is radius of circle is \[{{(x-1)}^{2}}+{{y}^{2}}={{r}^{2}}\] Since, circle is inscribed in ellipse \[\therefore \] Tangent of ellipse and tangent of circle at \[(h,k)\] are coincide. Now, slope of tangent of ellipse is \[2(x-1)+2y\frac{dy}{dx}=0\] \[\frac{dy}{dx}=\frac{-(x-1)}{y}\] \[{{\left( \frac{dy}{dx} \right)}_{(h,k)}}=-\left( \frac{h-1}{k} \right)\] Since, they are coincide \[\therefore \] \[\frac{-h}{4k}=\frac{-(h-1)}{k}\,\,\Rightarrow \,h=\frac{4}{3}\] Put the value of h in equation of ellipse, we get \[k=\frac{4\sqrt{2}}{3}\] \[\therefore \] \[{{r}^{2}}={{\left( \frac{4}{3}-1 \right)}^{2}}+{{\left( \frac{4\sqrt{2}}{3} \right)}^{2}}\] \[r=\sqrt{\frac{1}{9}+\frac{32}{9}}=\sqrt{\frac{33}{9}}=\sqrt{\frac{11}{3}}\]