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Question Number 123426 by MJS_new last updated on 25/Nov/20
find the radii of all circles with center C  tangenting the curve h  C= ((5),(1) )  h: x^2 −y^2 =1
$$\mathrm{find}\:\mathrm{the}\:\mathrm{radii}\:\mathrm{of}\:\mathrm{all}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{center}\:{C} \\ $$$$\mathrm{tangenting}\:\mathrm{the}\:\mathrm{curve}\:{h} \\ $$$${C}=\begin{pmatrix}{\mathrm{5}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${h}:\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{1} \\ $$
Commented by ajfour last updated on 25/Nov/20
Yes Sir four circles, thanks!
$${Yes}\:{Sir}\:{four}\:{circles},\:{thanks}! \\ $$
Commented by ajfour last updated on 25/Nov/20
Commented by MJS_new last updated on 25/Nov/20
I get 4. your green circle represents 2 very  close together
$$\mathrm{I}\:\mathrm{get}\:\mathrm{4}.\:\mathrm{your}\:{green}\:\mathrm{circle}\:\mathrm{represents}\:\mathrm{2}\:\mathrm{very} \\ $$$$\mathrm{close}\:\mathrm{together} \\ $$
Commented by MJS_new last updated on 25/Nov/20
I get  r_1 ≈2.70949170  r_2 ≈4.05595543  r_3 ≈4.16548440  r_4 ≈6.07096532  they are the positive solutions of  r^8 −78r^6 +2050r^4 −21762r^2 +77234=0
$$\mathrm{I}\:\mathrm{get} \\ $$$${r}_{\mathrm{1}} \approx\mathrm{2}.\mathrm{70949170} \\ $$$${r}_{\mathrm{2}} \approx\mathrm{4}.\mathrm{05595543} \\ $$$${r}_{\mathrm{3}} \approx\mathrm{4}.\mathrm{16548440} \\ $$$${r}_{\mathrm{4}} \approx\mathrm{6}.\mathrm{07096532} \\ $$$$\mathrm{they}\:\mathrm{are}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{solutions}\:\mathrm{of} \\ $$$${r}^{\mathrm{8}} −\mathrm{78}{r}^{\mathrm{6}} +\mathrm{2050}{r}^{\mathrm{4}} −\mathrm{21762}{r}^{\mathrm{2}} +\mathrm{77234}=\mathrm{0} \\ $$
Commented by MJS_new last updated on 25/Nov/20
sorry, corrected. it′s “green” not “yellow”
$$\mathrm{sorry},\:\mathrm{corrected}.\:\mathrm{it}'\mathrm{s}\:“\mathrm{green}''\:\mathrm{not}\:“\mathrm{yellow}'' \\ $$

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