maximum radius of a circle inscribed in an ellipse Consider an ellipse with major and minor axes of length $10$ and $8$ respectively. The radius of the largest circle that can be inscribed in this ellipse, given that the centre of this circle is one of the focus of the ellipse.
I attempted to solve this by using a very simple concept:
All points of the circle should either lie inside or on the circle. Hence assuming any point on the circle to be $x=ae+r\cos \theta$ and $y=r\sin \theta$  which satisfies:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}<1$$
I got a quadratic in $\cos \theta$ and I made the equation to be true independent of theta which gave $r\in [a-ae,a+ae]$ what is wrong to solve it by this method.
I would also like to know any other methods to solve this problem.
 A: Hint:


*

*Parametrize the ellipse as $(x,y) = (5 \cos t, 4 \sin t)$.

*Note that a focus is located at $(3,0)$.

*Obtain an expression for the distance of a point on the ellipse to this focus.

*Find a value of $t$ for which this distance is minimized.

*Compute the minimum distance. This is the radius of the largest circle centered at the focus.

A: Oh, this problem can be easily solved by using the basic properties of ellipse.
We want to find the radius of the largest circle centred at the focus of ellipse, which means we need to find the closest pt of the ellipse from the focus, and the distance between the focus this point will be the largest possible radius.
Note: There are 2 possibilities, that there may exist 2 points (above and below the major axis,symmetric) or just a single pt which is the end point of major axis. ( We will prove that only 1 of these is possible)
There are 2 ways to solve this-
1 ). The distance of any point on the ellipse from the focus in eccentricity times the distance of the point from its directrix, so the the shortest point from the directrix to the ellipse is the point on the major axis. ( Hence this is the pt, shortest from the focus also )
2 ). Let F1 and F2 be the focus of the ellipse. let F1 be the centre of the circle C and it touches the ellipse at P. Now consider a ray emanating from F1 and bouncing of at P on the tangent L1(say). Now as L1 is the tangent to the cirlce it must return to its centre (F1). Also L1 is tangent to ellipse this means it must pass theough F2.
The two case is possible iff PF1F2 are collinear.
So I conclude that the required radius of the circle is the distance between focus and the end pt of major axis which in other words is
a-ae . ( a is the semi major axis length and e is the eccentricity)
A: Instead of circle parametrization, write the ellipse in polar coordinates with focus-centred:
$$r=\frac{b^2}{a+c\cos \theta}$$
The maximal interior circle has its radius being the minimal focal distance (i.e. at apogee):
\begin{align}
  r_{-} &= \frac{b^2}{a+c} \\
  &= \frac{b^2}{a+\sqrt{a^2-b^2}} \\
  &= \frac{4^2}{5+\sqrt{5^2-4^2}} \\
  &= 2
\end{align}
A: Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$. 
Parametric point on the ellipse is $(a\cos t, b\sin t)$ . Distance of centre $(0,0)$ from this point is given by $(a-ae \cos t)$. (Substitute in distance formula and simplify to obtain that expression)
$(a-ae \cos t)$ in this case is $(5-3\cos t)$. 
It is minimum for $\cos t=1$ and its value is $2$. Hence , the radius of the largest circle that can be inscribed is $2$ units. 
