<s>[INST] Find radius of largest circle within ellipse $\frac{x^2}{9} + \frac{y^2}4 = 1$ with their intersection only at $(3,0)$ 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? 
 [/INST] 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 $3-a= \frac43$.
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