Question

The radius of the largest circle with centre $$\left( {1,0} \right)$$ that can be inscribed in the ellipse ${x^2} + 4{y^2} = 16$ is
(A) $\dfrac{{11}}{3}$
(B) $\dfrac{{11}}{{\sqrt 3 }}$
(C) $\dfrac{{\sqrt {11} }}{3}$
(D) $\sqrt {\dfrac{{11}}{3}} $

Answer 1

Hint: Let a general point on an ellipse which is of form $\left( {a\cos \theta , b\sin \theta } \right)$ and draw a normal on ellipse at that point. Since any normal of a circle always passes through its centre, by putting coordinates of centre in the equation of normal we can find the value of $\theta $ that gives the common point of ellipse and circle and the distance between this point and centre gives the radius of the largest circle that can be inscribed in an ellipse.

Complete step by step answer:
Here, Given equation of an ellipse is ${x^2} + 4{y^2} = 16$
This equation can also be written as $\eqalign{
  & \dfrac{{{x^2}}}{{16}} + \dfrac{{4{y^2}}}{{16}} = 1 \cr
  & \dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{4} = 1 \cr} $
Equivalent to $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
This gives ${a^2} = 16\,\,$and ${b^2} = 4$
$\therefore a = \pm 4\,$ and ${b^2} = \pm 2$
General point on an ellipse is given by $\left( {a\cos \theta ,b\sin \theta } \right)$
Thus, the point on ellipse is $\left( {4\cos \theta ,2\sin \theta } \right)$
Now, equation of normal on an ellipse is given by $ax\sec \theta - by\cos ec\theta = {a^2} - {b^2}$
Equation of normal at any arbitrary point on this ellipse is $4x\sec \theta - 2y\cos ec\theta = 16 - 4$
$ \Rightarrow 4x\sec \theta - 2y\cos ec\theta = 12$
Centre of circle $\left( {1,0} \right)$ satisfy this equation as normal passes through the centre so, $4\left( 1 \right)\sec \theta - 2\left( 0 \right)\cos ec\theta = 12$
$\Rightarrow 4\sec \theta = 12$
$\eqalign{
 & \Rightarrow \sec \theta = 3 \cr
 & \therefore \cos \theta = \dfrac{1}{3} \cr} $
$\sin \theta = \dfrac{{\sqrt {9 - 1} }}{3} = \dfrac{{\sqrt 8 }}{3} = \dfrac{{2\sqrt 2 }}{3}$
Thus, required point on ellipse is $\left( {\dfrac{4}{3},\dfrac{{4\sqrt 2 }}{3}} \right)$
Distance between two points is given by formula $ = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} $
Now, radius of circle is $\sqrt {{{\left( {\dfrac{4}{3} - 1} \right)}^2} + {{\left( {\dfrac{{4\sqrt 2 }}{3} - 0} \right)}^2}} $
$\eqalign{
  & = \sqrt {{{\left( {\dfrac{1}{3}} \right)}^2} + {{\left( {\dfrac{{4\sqrt 2 }}{3}} \right)}^2}} \cr
  & = \sqrt {\left( {\dfrac{1}{9}} \right) + \left( {\dfrac{{32}}{9}} \right)} \cr
  & = \sqrt {\dfrac{{33}}{9}} \cr
  & = \sqrt {\dfrac{{11}}{3}} \cr} $
Thus, the radius of the largest inscribed circle in ellipse is $\sqrt {\dfrac{{11}}{3}} $ units.

Hence option ‘D’ is correct.

Note:
Alternative methods: let the radius of the required circle be $r$and centre is$\left( {1,0} \right)$ so the equation of circle is ${\left( {x - 1} \right)^2} + {\left( {y - 0} \right)^2} = {r^2}$ and the given equation of ellipse is ${x^2} + 4{y^2} = 16$ . since circle is inscribed in the ellipse so they must have at least one common point. Now after eliminating one variable from these two equations we get a quadratic equation in another variable. Since this equation has equal and common roots as both curves intersect at the same point so discriminants of the given equation is zero. By solving this discriminants equation we get the value of radius of the required circle.