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If curve $$A(Z:|Z-3|+|Z+3|=8)$$ and $$B(Z:|Z-3|=K),K\in\mathbb{R^+}$$If $B$ touches the curve $'A'$ internally,and given another curve $$C(Z:||Z-3|-|Z+3||=4)$$The question is to prove that $A$ and $C$ intersect orthogonally and also prove that $B$ touches $C$ internally.

I had not encountered such questions before.I understand that $A$ represents a ellipse with length of major axis as $8$ and focii $(-3,0)$ and $(3,0)$.Similarly $B$ represents a circle and $C$ a hyperbola.$B$ and $C$ are having same focii.The hint that was provided to me was that two confocalconics always intersect orthogonally.However I had no reason to accept this and I couldnot prove the fact.Any help is appreciated.Thanks.

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This answer proves that if $a^2-b^2=\alpha^2+\beta^2$, then the ellipse $$ \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1 $$ and the hyperbola $$ \frac{x^2}{α^2} −\frac{y^2}{β^2} = 1 $$ intersect orthogonally.

If follows from this that $A$ and $C$ intersect orthogonally.

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Now since $B$ touches $A$ internally, we have to have $$K=4-3=1$$ (the $4$ comes from that $|4-3|+|4+3|=8$)

It follows from $3-2=K$ with $||2-3|-|2+3||=4$ that $B$ touches $C$ internally.

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    Sorry I forgot to include the modulus sign.Thanks for pointing out. I have edited the question – 2017-02-26
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    @navinstudent: I've edited the answer. – 2017-02-26