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Question Number 123080 by ajfour last updated on 22/Nov/20

Commented by ajfour last updated on 22/Nov/20

$F i n d r a d i i r a n d R .$
Commented by MJS_new last updated on 24/Nov/20

$we can select two points P = (\begin{matrix} p \\ p^{2} \end{matrix}) for the$ $circle touching the x - axis and Q = (\begin{matrix} q \\ q^{2} \end{matrix}) for$ $the circle touching the y - axis and get the$ $equations of the two circles with radii r_{p}, r_{q}$ $(in the 1^{st} quadrant) . then calculate the$ $distances of their centers to the origin d_{p}, d_{q} .$ $we get 2 equations which can be solved onlyo$ $approximately .$ $(1) r_{p} = r_{q}$ $(2) d_{p} = d_{q}$ $I get p \approx . 917194023 and q \approx 1.0856955$ $\Rightarrow r_{p} = r_{q} \approx . 568931542 and d_{p} = d_{q} \approx 1.52669097$ $\Rightarrow R \approx 2.09562251$
Commented by mr W last updated on 24/Nov/20

$a n s w e r i s c o r r e c t!$
	Answered by mr W last updated on 24/Nov/20

	$P (p, p^{2})$ $A (\sqrt{{(R - r)}^{2} - r^{2}}, r)$ $\frac{p^{2} - r}{p - \sqrt{R (R - 2 r)}} = - \frac{1}{2 p}$ $\Rightarrow 2 p^{3} - (2 r - 1) p - \sqrt{R (R - 2 r)} = 0 . . . (i)$ ${(p - \sqrt{R (R - 2 r)})}^{2} + {(p^{2} - r)}^{2} = r^{2}$ $\Rightarrow p^{4} - (2 r - 1) p^{2} - 2 \sqrt{R (R - 2 r)} p + R (R - 2 r) = 0 . . (i i)$ $(i) \times p - (i i) \times 2 :$ $(2 r - 1) p^{2} + 3 \sqrt{R (R - 2 r)} p - 2 R (R - 2 r) = 0$ $\Rightarrow p = \frac{\sqrt{R (R - 2 r)} (\sqrt{16 r + 1} - 3)}{2 (2 r - 1)}$ $p u t i t i n t o (i) :$ $R^{2} - 2 r R - 2 (\sqrt{16 r + 1} - 1) {(\frac{2 r - 1}{\sqrt{16 r + 1} - 3})}^{3} = 0$ $\Rightarrow R = r + \sqrt{r^{2} + 2 (\sqrt{16 r + 1} - 1) {(\frac{2 r - 1}{\sqrt{16 r + 1} - 3})}^{3}} . . . (I)$ $Q (q, q^{2})$ $B (r, \sqrt{R (R - 2 r)})$ $\frac{q^{2} - \sqrt{R (R - 2 r)}}{q - r} = - \frac{1}{2 q}$ $2 q^{3} - [2 \sqrt{R (R - 2 r)} - 1] q - r = 0 . . . (i i i)$ ${(q^{2} - \sqrt{R (R - 2 r)})}^{2} + {(q - r)}^{2} = r^{2}$ $q^{4} - [2 \sqrt{R (R - 2 r)} - 1] q^{2} - 2 r q + R (R - 2 r) = 0 . . . (i v)$ $(i i i) \times p - (i v) \times 2 :$ $[2 \sqrt{R (R - 2 r)} - 1] q^{2} + 3 r q - 2 R (R - 2 r) = 0$ $\Rightarrow q = \frac{\sqrt{9 r^{2} + 8 R (R - 2 r) [2 \sqrt{R (R - 2 r)} - 1]} - 3 r}{2 [2 \sqrt{R (R - 2 r)} - 1]}$ $p u t i t i n t o (i i i) :$ $\Rightarrow {[\frac{\sqrt{9 r^{2} + 8 R (R - 2 r) [2 \sqrt{R (R - 2 r)} - 1]} - 3 r}{2 \sqrt{R (R - 2 r)} - 1}]}^{3} -$ $2 \sqrt{9 r^{2} + 8 R (R - 2 r) [2 \sqrt{R (R - 2 r)} - 1]} + 2 r = 0 . . . (I I)$ $p u t (I) i n t o (I I) w e g e t a n e q u a t i o n$ $f o r r w h i c h g i v e s r \approx 0.5689 . p u t t h i s$ $i n t o (I) w e g e t R \approx 2.0955 .$
	Commented by mr W last updated on 24/Nov/20

	Commented by ajfour last updated on 25/Nov/20

	$T h a n k y o u M j S S i r a n d m r W S i r!$
	Answered by ajfour last updated on 26/Nov/20

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