An ellipse is drawn with the major and minor axes of lengths of $10$ and $8$ respectively
Using one focus as center is drawn such that it is tangent to the ellipse.
The radius of the circle is
$\bf{My\; Try::}$ Here I have assume Center of ellipse is at origin.
So equation of ellipse is $\displaystyle \frac{x^2}{25}+\frac{y^2}{16}=1$
Using Some calculation $\displaystyle e = \frac{3}{5}$ and So focus is $\displaystyle (\pm 3,0)$
Now equation of Circle whose one center is at one focus $(3,0)$
So equation of Circle is $(x-3)^2+y^2=r^2\;,$ Where radius is $=r$
Now eliminating $y$ from these two equation, We get $\displaystyle \frac{x^2}{25}+\frac{r^2-(x-3)^2}{16} = 1$
So we get $\displaystyle 16x^2+25r^2-25(x-3)^2=400\Rightarrow 16x^2+25r^2-25x^2-225+150x=400$
So $\displaystyle -9x^2+150x+25r^2-625=0\Rightarrow 9x^2-150x-25(r^2-25)=0$
Now given Ellipse and Circle are Tangent.
So we will put $\bf{Discriminant =0}$
So $\displaystyle (150)^2-+4\cdot 25 \cdot 9 (r^2-25) =0\Rightarrow r^2=0\Rightarrow r=0$
But When I solved Using Parametric Coordinate I get $r=2$
I did not understand Why I am not getting Same answer , plz explain me
Thanks
Best Answer
The circles of centers $(3,0)$ and radius $r = 5 \pm 3$ are clearly tangent to the ellipse. So why aren't the discriminant of your equation in $x$ equal to $0$ for these values of $x$? Because there are also other solutions in $x$. The problem is that the corresponding $y$'s are complex ( try solving the system obtained).
One way to find tangent circles to an ellipse is using Lagrange multipliers. You want to find level curves of the function $f(x,y) = (x-3)^2 + y^2$ that are tangent to the curve $g(x,y) = x^2/25 + y^2/16-1=0$. So get the Lagrange function $F\colon = f - t \cdot g$ and solve the system in $(x,y,t)$ $\colon \frac{\partial F}{\partial l}=0$, $l=x,y,t$. You will find in fact three circles, the two obvious ones and (surprise!) the one of radius $r=0$ ( although for the last one the points where the curves are tangent have complex coordinates).
$\bf{ Added}$: Let's see how that Lagrange system works out:
$$ 2( x-3) - t \frac{2 x}{25} =0 \\ 2 y - t \frac{y}{8} = 0\\ x^2/25 + y^2/16 = 1$$
The second equation is $y(16-t) = 0$. Hence $y=0$ or $t = 16$. In the first case $y=0$ and so $x = \pm 5$. In the second case, $t=16$ and from the first equation we get $x=\frac{25}{3}$, and from the third, $y=\pm \frac{16 i}{3}$. Note that $(x-3)^2 + y^2 =( \frac{16}{3})^2 + ( \frac{16i}{3})^2=0$.
Let's also see what happens with the circle of radius $2$ with center $(3,0)$. The intersection of the circle with the ellipse is given by the system $$(x-3)^2 + y^2 = 4 \\ x^2/25 + y^2/16 = 1$$ with solutions $(3,0)$ ( with multiplicity $2$ - but not for the equation in $x$ but as the fact that the curves are tangent there ) and two more complex solutions $(x,y) = (35/3, \pm \frac{16 \sqrt{10} i}{3})$