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Question Number 43783 by ajfour last updated on 15/Sep/18

Commented by ajfour last updated on 16/Sep/18

$c a n a n a p p p l o t i t i f i t b e u n i q u e ?$
Commented by ajfour last updated on 15/Sep/18

$S h o u l d i t b e u n i q u e ?$
Commented by MrW3 last updated on 15/Sep/18

$I {d o n}^{'} t t h i n k {i t}^{'} s u n i q u e .$
Commented by ajfour last updated on 15/Sep/18

$i h a v e a n o t i o n, {i t}^{'} l l b e a s i n g l e$ $e q u a t i o n .$
Commented by MJS last updated on 16/Sep/18

$parabola \cap circle has got 4 solutions, real$ $or complex \Rightarrow {we}^{'} ll get polynomes of 4^{th} degree$ $including 3 unknown and 1 parameter$ $we have 6 unknown p, q, x_{A}, y_{A}, x_{B}, y_{B}$ $par : y = q - {p x}^{2}$ $par \cap red circle = A = (\begin{matrix} x_{A} \\ y_{A} \end{matrix})$ $par \cap blue circle = B = (\begin{matrix} x_{B} \\ y_{B} \end{matrix})$ $but I {can}^{'} t see more than 2, maybe 4 equations$ $looks bad . . . but I^{'} m usually too pessimistic$
	Answered by ajfour last updated on 17/Sep/18

	$l e t p a r a b o l a e q . b e$ $y = c - {A x}^{2}$ $l e t p o i n t o f t a n g e n c y w i t h c i r c l e$ $w i t h c e n t r e (r, r) a n d r a d i u s r b e$ $P (h, c - {A h}^{2})$ $s l o p e o f t a n g e n t t o p a r a b o l a =$ $s l o p e o f t a n g e n t t o c i r c l e$ $\Rightarrow - 2 A h = \frac{c - {A h}^{2} - r}{h - r}$ $A l s o$ ${(h - r)}^{2} + {(c - {A h}^{2} - r)}^{2} = r^{2}$ $\Rightarrow {(h - r)}^{2} + \frac{{(h - r)}^{2}}{4 A^{2} h^{2}} = r^{2}$ $\Rightarrow {(h - r)}^{2} = 4 A^{2} h^{3} (2 r - h)$ $. . . . . .$
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