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

Commented by ajfour last updated on 22/Nov/20

Find radii r and R.

$${Find}\:{radii}\:\:{r}\:{and}\:{R}. \\ $$

Commented by MJS_new last updated on 24/Nov/20

we can select two points P= ((p),(p^2 ) ) for the circle touching the x−axis and Q= ((q),(q^2 ) ) 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≈.917194023 and q≈1.0856955 ⇒ r_p =r_q ≈.568931542 and d_p =d_q ≈1.52669097 ⇒ R≈2.09562251

$$\mathrm{we}\:\mathrm{can}\:\mathrm{select}\:\mathrm{two}\:\mathrm{points}\:{P}=\begin{pmatrix}{{p}}\\{{p}^{\mathrm{2}} }\end{pmatrix}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\mathrm{touching}\:\mathrm{the}\:{x}−\mathrm{axis}\:\mathrm{and}\:{Q}=\begin{pmatrix}{{q}}\\{{q}^{\mathrm{2}} }\end{pmatrix}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{circle}\:\mathrm{touching}\:\mathrm{the}\:{y}−\mathrm{axis}\:\mathrm{and}\:\mathrm{get}\:\mathrm{the} \\ $$$$\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{radii}\:{r}_{{p}} ,\:{r}_{{q}} \\ $$$$\left(\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{quadrant}\right).\:\mathrm{then}\:\mathrm{calculate}\:\mathrm{the} \\ $$$$\mathrm{distances}\:\mathrm{of}\:\mathrm{their}\:\mathrm{centers}\:\mathrm{to}\:\mathrm{the}\:\mathrm{origin}\:{d}_{{p}} ,\:{d}_{{q}} . \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{2}\:\mathrm{equations}\:\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{onlyo} \\ $$$$\mathrm{approximately}. \\ $$$$\left(\mathrm{1}\right)\:{r}_{{p}} ={r}_{{q}} \\ $$$$\left(\mathrm{2}\right)\:{d}_{{p}} ={d}_{{q}} \\ $$$$\mathrm{I}\:\mathrm{get}\:{p}\approx.\mathrm{917194023}\:\mathrm{and}\:{q}\approx\mathrm{1}.\mathrm{0856955} \\ $$$$\Rightarrow\:{r}_{{p}} ={r}_{{q}} \approx.\mathrm{568931542}\:\mathrm{and}\:{d}_{{p}} ={d}_{{q}} \approx\mathrm{1}.\mathrm{52669097} \\ $$$$\Rightarrow\:{R}\approx\mathrm{2}.\mathrm{09562251} \\ $$

Commented by mr W last updated on 24/Nov/20

answer is correct!

$${answer}\:{is}\:{correct}! \\ $$

Answered by mr W last updated on 24/Nov/20

P(p,p^2 ) A((√((R−r)^2 −r^2 )),r) ((p^2 −r)/(p−(√(R(R−2r)))))=−(1/(2p)) ⇒2p^3 −(2r−1)p−(√(R(R−2r)))=0 ...(i) (p−(√(R(R−2r))))^2 +(p^2 −r)^2 =r^2 ⇒p^4 −(2r−1)p^2 −2(√(R(R−2r)))p+R(R−2r)=0 ..(ii) (i)×p−(ii)×2: (2r−1)p^2 +3(√(R(R−2r)))p−2R(R−2r)=0 ⇒p=(((√(R(R−2r)))((√(16r+1))−3))/(2(2r−1))) put it into (i): R^2 −2rR−2((√(16r+1))−1)(((2r−1)/( (√(16r+1))−3)))^3 =0 ⇒R=r+(√(r^2 +2((√(16r+1))−1)(((2r−1)/( (√(16r+1))−3)))^3 )) ...(I) Q(q,q^2 ) B(r,(√(R(R−2r)))) ((q^2 −(√(R(R−2r))))/(q−r))=−(1/(2q)) 2q^3 −[2(√(R(R−2r)))−1]q−r=0 ...(iii) (q^2 −(√(R(R−2r))))^2 +(q−r)^2 =r^2 q^4 −[2(√(R(R−2r)))−1]q^2 −2rq+R(R−2r)=0 ...(iv) (iii)×p−(iv)×2: [2(√(R(R−2r)))−1]q^2 +3rq−2R(R−2r)=0 ⇒q=(((√(9r^2 +8R(R−2r)[2(√(R(R−2r)))−1]))−3r)/(2[2(√(R(R−2r)))−1])) put it into (iii): ⇒[(((√(9r^2 +8R(R−2r)[2(√(R(R−2r)))−1]))−3r)/(2(√(R(R−2r)))−1))]^3 − 2(√(9r^2 +8R(R−2r)[2(√(R(R−2r)))−1]))+2r=0 ...(II) put (I) into (II) we get an equation for r which gives r≈0.5689. put this into (I) we get R≈2.0955.

$${P}\left({p},{p}^{\mathrm{2}} \right) \\ $$$${A}\left(\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} },{r}\right) \\ $$$$\frac{{p}^{\mathrm{2}} −{r}}{{p}−\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}}=−\frac{\mathrm{1}}{\mathrm{2}{p}} \\ $$$$\Rightarrow\mathrm{2}{p}^{\mathrm{3}} −\left(\mathrm{2}{r}−\mathrm{1}\right){p}−\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}=\mathrm{0}\:\:\:...\left({i}\right) \\ $$$$\left({p}−\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}\right)^{\mathrm{2}} +\left({p}^{\mathrm{2}} −{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow{p}^{\mathrm{4}} −\left(\mathrm{2}{r}−\mathrm{1}\right){p}^{\mathrm{2}} −\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}{p}+{R}\left({R}−\mathrm{2}{r}\right)=\mathrm{0}\:\:\:..\left({ii}\right) \\ $$$$\left({i}\right)×{p}−\left({ii}\right)×\mathrm{2}: \\ $$$$\left(\mathrm{2}{r}−\mathrm{1}\right){p}^{\mathrm{2}} +\mathrm{3}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}{p}−\mathrm{2}{R}\left({R}−\mathrm{2}{r}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}=\frac{\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}\left(\sqrt{\mathrm{16}{r}+\mathrm{1}}−\mathrm{3}\right)}{\mathrm{2}\left(\mathrm{2}{r}−\mathrm{1}\right)} \\ $$$${put}\:{it}\:{into}\:\left({i}\right): \\ $$$${R}^{\mathrm{2}} −\mathrm{2}{rR}−\mathrm{2}\left(\sqrt{\mathrm{16}{r}+\mathrm{1}}−\mathrm{1}\right)\left(\frac{\mathrm{2}{r}−\mathrm{1}}{\:\sqrt{\mathrm{16}{r}+\mathrm{1}}−\mathrm{3}}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow{R}={r}+\sqrt{{r}^{\mathrm{2}} +\mathrm{2}\left(\sqrt{\mathrm{16}{r}+\mathrm{1}}−\mathrm{1}\right)\left(\frac{\mathrm{2}{r}−\mathrm{1}}{\:\sqrt{\mathrm{16}{r}+\mathrm{1}}−\mathrm{3}}\right)^{\mathrm{3}} }\:\:\:\:\:...\left({I}\right) \\ $$$$ \\ $$$${Q}\left({q},{q}^{\mathrm{2}} \right) \\ $$$${B}\left({r},\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}\right) \\ $$$$\frac{{q}^{\mathrm{2}} −\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}}{{q}−{r}}=−\frac{\mathrm{1}}{\mathrm{2}{q}} \\ $$$$\mathrm{2}{q}^{\mathrm{3}} −\left[\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−\mathrm{1}\right]{q}−{r}=\mathrm{0}\:\:\:...\left({iii}\right) \\ $$$$ \\ $$$$\left({q}^{\mathrm{2}} −\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}\right)^{\mathrm{2}} +\left({q}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${q}^{\mathrm{4}} −\left[\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−\mathrm{1}\right]{q}^{\mathrm{2}} −\mathrm{2}{rq}+{R}\left({R}−\mathrm{2}{r}\right)=\mathrm{0}\:\:\:...\left({iv}\right) \\ $$$$\left({iii}\right)×{p}−\left({iv}\right)×\mathrm{2}: \\ $$$$\left[\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−\mathrm{1}\right]{q}^{\mathrm{2}} +\mathrm{3}{rq}−\mathrm{2}{R}\left({R}−\mathrm{2}{r}\right)=\mathrm{0} \\ $$$$\Rightarrow{q}=\frac{\sqrt{\mathrm{9}{r}^{\mathrm{2}} +\mathrm{8}{R}\left({R}−\mathrm{2}{r}\right)\left[\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−\mathrm{1}\right]}−\mathrm{3}{r}}{\mathrm{2}\left[\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−\mathrm{1}\right]} \\ $$$${put}\:{it}\:{into}\:\left({iii}\right): \\ $$$$\Rightarrow\left[\frac{\sqrt{\mathrm{9}{r}^{\mathrm{2}} +\mathrm{8}{R}\left({R}−\mathrm{2}{r}\right)\left[\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−\mathrm{1}\right]}−\mathrm{3}{r}}{\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−\mathrm{1}}\right]^{\mathrm{3}} − \\ $$$$\:\:\:\:\:\mathrm{2}\sqrt{\mathrm{9}{r}^{\mathrm{2}} +\mathrm{8}{R}\left({R}−\mathrm{2}{r}\right)\left[\mathrm{2}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−\mathrm{1}\right]}+\mathrm{2}{r}=\mathrm{0}\:\:\:...\left({II}\right) \\ $$$$ \\ $$$${put}\:\left({I}\right)\:{into}\:\left({II}\right)\:{we}\:{get}\:{an}\:{equation} \\ $$$${for}\:{r}\:{which}\:{gives}\:{r}\approx\mathrm{0}.\mathrm{5689}.\:{put}\:{this} \\ $$$${into}\:\left({I}\right)\:{we}\:{get}\:{R}\approx\mathrm{2}.\mathrm{0955}. \\ $$

Commented by mr W last updated on 24/Nov/20

Commented by ajfour last updated on 25/Nov/20

Thank you MjS Sir and mrW Sir!

$${Thank}\:{you}\:{MjS}\:{Sir}\:{and}\:{mrW}\:{Sir}! \\ $$

Answered by ajfour last updated on 26/Nov/20

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