if a curve $A=\{z : |z-3|+|z+3|=8\}$ and $B=\{z : |z-3|=k\},\; k\in \mathbb R^+$ touch the curve $A$ internally, then which are correct

Question

if a curve $A=\{Z : |Z-3|+|Z+3|=8\}$ and $B=\{Z : |Z-3|=k\},\; k\in \mathbb R^+$ touches the curve $A$ internally, and given another curve $C = \bigl\{Z : \bigl||Z-3|-|Z+3|\bigr|=4\bigr\},\;$ where $Z$ denotes complex number $x+iy$. Then

(A) $A$ and $C$ intersect orthogonally and $k=1$'

(B) $k=1$

(C) $k=\dfrac{3}{2}$

(D) $B$ touches $C$ internally.

My Approach:

Curve $A$ is ellipse given by $\dfrac{x^2}{16}+\dfrac{y^2}{7}=1\;$ and curve $B$ is circle given by $(x-3)^2+y^2=k^2$.

Since both curve are touching each other so i solved both curve together and Put discriminant equal to $0$

Steps: $\dfrac{x^2}{16}+\dfrac{k^2-(x-3)^2}{7}=1$

$\implies\;\;7x^2+16(k^2-(x-3)^2)=112$

$\implies\;\; 9x^2-96x+(256-16k^2)=0$

Now I made discriminant $D=0$ because both curve are touching each other.

Above equation give me $k=0$.

But after verifying from Desmos $k=1$ is solution and i know how to do it graphically.

My Doubt:

$1.$ What is going wrong in my method

This problem is same as A problem on confocal conics but in this problem OP's Doubt is different.

Answer 1

This is a slightly detailed explanation of the answer given by @aschepler in the comments:

Let the set of points in $\mathbb{R}^2$ satisfying $\frac{x^2}{16} + \frac{y^2}{7} = 1$ be denoted by $E$, let the set of points in $\mathbb{R}^2$ satisfying $(x-3)^2 + y^2 = k$ be $C$, and also let the set of points in $\mathbb{R}^2$ satisfying $\frac{x^2}{16} + \frac{k^2 - (x-3)^2}{7} = 1$ be denoted by $T$.

Now suppose you have given that $C$ touches $E$ internally. Assuming that $C$ and $E$ have no other points of intersection, it implies that $C \cap E$ has exactly one point in $\mathbb{R}^2$. However $C \cap E \neq T$, rather $(x,y) \in C \cap E \Rightarrow (x,y) \in T$, which means that $C \cap E \subseteq T$.
So the fact that any equation representing $C \cap E$ must have exactly one distinct real root, does not imply that the equation representing $T$ must have one real root. So you can't set the discriminant of the quadratic to zero, because that quadratic may have more than one root.

What is the way to solve it then?

Let $C \cap E$ be the set of points in $\mathbb{R}^2$ satisfying $\frac{x^2}{16} + \frac{y^2}{7} = 1 \land y^2 = k^2 - (x-3)^2$. Now suppose $x_1(k)$ and $x_2(k)$ are roots of $\frac{x^2}{16} + \frac{k^2 - (x-3)^2}{7} = 1$, and their corresponding $y$ values got from $y^2 = k^2 - (x-3)^2$ be $\pm y_1(k)$ and $\pm y_2(k)$. So your possible solutions to $T'$ are $(x_1(k), \pm y_1(k))$ and $(x_2(k), \pm y_2(k))$. Now substituting them back in $\frac{x^2}{16} + \frac{y^2}{7} = 1$, for each of the 4 points you will get a value of $k$ for which that point is a solution of $C \cap E$.

This method is quite ugly and doesn't generalize very well, but I'm curious to see a generic criterion for exactly one point of intersection of two conics.

Answer 2

Suppose $P(x_0,y_0)$ is the intersection of $A$ and $B$. Suppose $y_0\neq0$. Then the slope of the tangent of $A$ at $P$ is $$ m_A=-\frac{7x_0}{16y_0} $$ and the slope of the tangent of $B$ at $P$ is $$ m_B=-\frac{x_0-3}{y_0}. $$ Since $A$ and $B$ touch at $P$, one has $$ -\frac{7x_0}{16y_0}=-\frac{x_0-3}{y_0} $$ which gives $$ x_0=\frac{16}{3}. $$ But $P_0$ is on $A$ which implies $$ \frac{x_0^2}{16}+\frac{y_0^2}{7}=1. $$ This equation doesn't have real solution. Thus $y_0=0$ and hence $x_0=4$ and $k=1$.