# 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.

Hint:

1. Parametrize the ellipse as $(x,y) = (5 \cos t, 4 \sin t)$.
2. Note that a focus is located at $(3,0)$.
3. Obtain an expression for the distance of a point on the ellipse to this focus.
4. Find a value of $t$ for which this distance is minimized.
5. Compute the minimum distance. This is the radius of the largest circle centered at the focus.

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}

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.