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Question Number 88656 by ajfour last updated on 12/Apr/20

Commented by ajfour last updated on 12/Apr/20

$I f A B = B C, f i n d r a d i u s r .$
	Answered by ajfour last updated on 12/Apr/20
	$(x−r)^2 +y^2 =r^2 circle y=x^2 parabola (x−r)^2 +x^4 =r^2 ⇒ x^3 +x−2r=0 (besides x=0) say real root is x=a. ⇒ a^3 +a−2r=0 .....(i) eq. of line: y=−((a^2 (x−2r))/((2r−a))) 2r+x_A =2a y_A =x_A ^2 = 4(r−a)^2 y_B = (y_A /2)=2(r−a)^2 =a^2 ⇒ a^2 −4ar+2r^2 =0 ....(ii) & a^3 +a−2r=0 ....(i) (i)−[a×(ii)] gives a+4a^2 r−2r−2ar^2 =0 ⇒ a^2 −((r/2)−(1/(4r)))a−(1/2)=0 a=((r/4)−(1/(8r)))+(√(((r/4)−(1/(8r)))^2 +(1/2))) from (ii): (2r−a)^2 =2r^2 ⇒ {((7r)/4)+(1/(8r))−(√(((r/4)−(1/(8r)))^2 +(1/2))) }^2 = 2r^2 ⇒ r≈ 2.65246$
	${(x - r)}^{2} + y^{2} = r^{2} c i r c l e$ $y = x^{2} p a r a b o l a$ ${(x - r)}^{2} + x^{4} = r^{2}$ $\Rightarrow x^{3} + x - 2 r = 0 (b e s i d e s x = 0)$ $s a y r e a l r o o t i s x = a .$ $\Rightarrow a^{3} + a - 2 r = 0 . . . . . (i)$ $e q . o f l i n e : y = - \frac{a^{2} (x - 2 r)}{(2 r - a)}$ $2 r + x_{A} = 2 a$ $y_{A} = x_{A}^{2} = 4 {(r - a)}^{2}$ $y_{B} = \frac{y_{A}}{2} = 2 {(r - a)}^{2} = a^{2}$ $\Rightarrow a^{2} - 4 a r + 2 r^{2} = 0 . . . . (i i)$ $& a^{3} + a - 2 r = 0 . . . . (i)$ $(i) - [a \times (i i)] g i v e s$ $a + 4 a^{2} r - 2 r - 2 {a r}^{2} = 0$ $\Rightarrow a^{2} - (\frac{r}{2} - \frac{1}{4 r}) a - \frac{1}{2} = 0$ $a = (\frac{r}{4} - \frac{1}{8 r}) + \sqrt{{(\frac{r}{4} - \frac{1}{8 r})}^{2} + \frac{1}{2}}$ $f r o m (i i) : {(2 r - a)}^{2} = 2 r^{2}$ $\Rightarrow {\frac{7 r}{4} + \frac{1}{8 r} - \sqrt{{(\frac{r}{4} - \frac{1}{8 r})}^{2} + \frac{1}{2}}}^{2}$ $= 2 r^{2}$ $\Rightarrow r \approx 2.65246$
	Commented by ajfour last updated on 12/Apr/20

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