Radius of largest circle in an ellipse

Question

Consider an ellipse whose major and minor axis of length $10$ and $8$ unit respectively. Then the radius of the largest circle which can be inscribed in such an ellipse if the circle's center is one focus of the ellipse.

What I tried:

Assuming that major axis and minor axis of an ellipse along coordinate axis.($x$ and $y$ axis respectively)

Then equation of ellipse is $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ where $a=5,b=4$

Then $\displaystyle b^2=a^2(1-e^2)$ . getting $\displaystyle e=3/5.$

coordinate of focus is $(\pm ae,0)=(\pm 3,0)$

Let equation of circle is $(x\pm 3)^2+y^2=r^2$

How do I solve it from here?

Answer 1

Denote $F, F'$ the foci of the ellipse. Assume we want to find the greatest circle inscribed in the ellipse, with center $F.$

• The circle and the ellipse have a common tangent line and also a common normal in their point (points) of tangency $P,$ eventually also $P'$ if they are two.

• The normal of an ellipse bisects the angle between the lines to the foci. In our case, the normal at $P$ bisects the angle between $PF$ and $PF'.$

• The normal of a circle passes through its center.

From the above properties follows that $P,F,F'$ are collinear. Therefore, $P$ is the vertex of the ellipse lying at the major axis nearest to $F,$ and the equation of the circle is $$(x-3)^2+y^2=|PF|^2=4$$

Answer 2

Try to obtain a $r$ such that the circle intersects with the ellipse at exactly 2 points. This is when the circle is tangent to the ellipse.

Answer 3

Eliminate $y$. Then find the value of $x$ where $r$ is minimised.

Answer 4

You want to minimize $r^2=(x-3)^2+y^2$ subject to $\frac{x^2}{25}+\frac{y^2}{16}=1$: $$L(x,y,\lambda)=(x-3)^2+y^2+\lambda\left(1-\frac{x^2}{25}-\frac{y^2}{16}\right)\\ \begin{cases}L_x=2x-6-\frac{2x\lambda}{25}=0\\ L_y=2y-\frac{y\lambda}{8}=0\\ L_{\lambda}=1-\frac{x^2}{25}-\frac{y^2}{16}=0\end{cases} \Rightarrow (x,y)=(\pm5,0);\\ r^2(5,0)=2^2 \ \text{(min)}$$ Notes:

1) WolframAlpha answer.

2) If you minimize $r^2=(x+3)^2+y^2$ s.t. $\frac{x^2}{25}+\frac{y^2}{16}=1$, you will get $r^2(-5,0)=2^2$.

3) Foci $(\pm c,0)$ can be found from $c^2=a^2-b^2=5^2-4^2=3^2 \Rightarrow c=\pm 3$.