Find radius of largest circle within ellipse $\frac{x^2}{9} + \frac{y^2}4 = 1$ with their intersection only at $(3,0)$

Question

An ellipse is defined by the equation $$\frac{x^2}{9} + \frac{y^2}4 = 1$$

Compute the radius of the largest circle that is internally tangent to the ellipse at $(3,0),$ and intersects the ellipse only at $(3,0).$

How can I write an equation for the largest circle within this ellipse if its equation is given? Is there a property or theorem I'm missing? Is there another way to solve?

Answer 1

Let the equation of the circle $(x-a)^2+y^2 =(3-a)^2$ that passes the point $(3,0)$. Then, substitute $y^2$ into $\frac{x^2}{9} + \frac{y^2}4 = 1$ to get

$$\frac59 x^2 -2ax +6a-5=0$$

Since the two shapes has only one common point $(3,0)$, the discriminate of above quadratic equation is zero, which yields $a= \frac53$ and hence the radius $3-a= \frac43$.

Answer 2

$$\frac{x^2}{9}+\frac{y^2}{4}=1\implies y'=-\frac{4x}{9y}\implies y''=\frac{4(xy'-y)}{9y^2}$$ The radius of curvature, R i.e. radius of circular arc that best approximates the given ellipse at the point $(3,0)$ will be the radius of largest circle which internally touches the ellipse at point $(3,0)$ only, is given as $$R=\left|\frac{(1+y'^2)^{3/2}}{y''}\right|=\left|\frac{\left(16x^2+81y^2\right)^{3/2}}{36(4x^2+9y^2)}\right|$$ Substituting $x=3$ & $y=0$ in above equation, radius $R$ of the largest circle touching ellipse internally at a single point $(3,0)$, is given as $$\color{blue}{R}=\left|\frac{\left(16(3)^2+81(0)^2\right)^{3/2}}{36(4(3)^2+9(0)^2)}\right|=\color{blue}{\frac43}$$