Question Number 69681 by ajfour last updated on 26/Sep/19
Commented by ajfour last updated on 26/Sep/19
$${Q}.\mathrm{69496}\:\:\:\left({A}\:{try}..\right) \\ $$
Commented by ajfour last updated on 26/Sep/19
$${Radius}\:{of}\:{circle}. \\ $$
Answered by ajfour last updated on 30/Sep/19
Commented by ajfour last updated on 30/Sep/19
![let cos α=a, cos β=h, cos γ=m, cos δ=u & sin α=b, sin β=k, sin γ=n, sin δ=v x_D =−qa+rm=ph−su ⇒ su+rm = qa+ph ...(i) y_D =qb+rn=pk+sv ⇒ sv−rn = qb−pk ...(ii) ∣CP ∣=R =∣CT∣ ⇒ _________________________ (−qa−m)^2 +(qb−n−R)^2 =R^2 [−qa+(r+6)m]^2 +(qb+(r+6)n−R]^2 =R^2 ⇒ r+6 and −1 are roots of eq. (−qa+mx)^2 +(qb+nx−R)^2 =R^2 (m^2 +n^2 )x^2 +2x[−qam+n(qb−R)] +q^2 a^2 +q^2 b^2 +R^2 −2qbR−R^2 =0 ⇒ x^2 +2x[−qam+n(qb−R)] +q^2 −2qbR=0 _________________________ ⇒ r+5=qam−n(qb−R) & m^2 +n^2 =1 .....(iii),(iv) [r+5+n(qb−R)]^2 =q^2 a^2 (1−n^2 ) ⇒ n^2 {(qb−R)^2 +q^2 a^2 }+ +2(r+5)(qb−R)n+(r+5)^2 −q^2 a^2 =0 _________________________ ∣CM∣=R =∣CS∣ ⇒ (ph+3u)^2 +(pk−3v−R)^2 =R^2 [ph−(s+5)u]^2 +[pk+(s+5)v−R]^2 =R^2 ⇒ s+5 and −3 are roots of eq. (ph−ux)^2 +(pk+vx−R)^2 =R^2 _________________________ ⇒ sum of roots (s+5)+(−3)= s+2=phu−v(pk−R) (as h^2 +k^2 =1) & u^2 +v^2 =1 ......(v),(vi) _________________________ plus _________________________ a^2 +b^2 =1 , h^2 +k^2 =1 ..(vii)&(viii) 2q=r+6 ....(ix) 4p=3(s+5) ....(x) 5(s+3)=6(r+1) ....(xi) further 2bR=q+2 & ..(xii) 2kR=p+4 ....(xiii) Thirteen eqs. in thirteen unknowns a,b,h,k,m,n,u,v,p,q,r,s,&R.](https://www.tinkutara.com/question/Q70067.png)
$${let}\:\mathrm{cos}\:\alpha={a},\:\mathrm{cos}\:\beta={h},\:\mathrm{cos}\:\gamma={m}, \\ $$$$\mathrm{cos}\:\delta={u}\:\:\:\&\:\:\mathrm{sin}\:\alpha={b},\:\mathrm{sin}\:\beta={k}, \\ $$$$\mathrm{sin}\:\gamma={n},\:\mathrm{sin}\:\delta={v} \\ $$$${x}_{{D}} =−{qa}+{rm}={ph}−{su} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{su}}+\boldsymbol{{rm}}\:=\:\boldsymbol{{qa}}+\boldsymbol{{ph}}\:\:\:\:\:…\left({i}\right) \\ $$$${y}_{{D}} ={qb}+{rn}={pk}+{sv} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{sv}}−\boldsymbol{{rn}}\:=\:\boldsymbol{{qb}}−\boldsymbol{{pk}}\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\mid{CP}\:\mid={R}\:=\mid{CT}\mid\:\Rightarrow \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\left(−{qa}−{m}\right)^{\mathrm{2}} +\left({qb}−{n}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\:\:\left[−{qa}+\left({r}+\mathrm{6}\right){m}\right]^{\mathrm{2}} +\left({qb}+\left({r}+\mathrm{6}\right){n}−{R}\right]^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\:{r}+\mathrm{6}\:{and}\:−\mathrm{1}\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\left(−{qa}+{mx}\right)^{\mathrm{2}} +\left({qb}+{nx}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$$$\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}{x}\left[−{qam}+{n}\left({qb}−{R}\right)\right] \\ $$$$\:+{q}^{\mathrm{2}} {a}^{\mathrm{2}} +{q}^{\mathrm{2}} {b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{qbR}−{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +\mathrm{2}{x}\left[−{qam}+{n}\left({qb}−{R}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{q}^{\mathrm{2}} −\mathrm{2}{qbR}=\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\:\:{r}+\mathrm{5}={qam}−{n}\left({qb}−{R}\right)\:\: \\ $$$$\&\:\:\:\:\:\:\:\:{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{1}\:\:\:\:\:\:\:\:…..\left({iii}\right),\left({iv}\right) \\ $$$$\:\left[{r}+\mathrm{5}+{n}\left({qb}−{R}\right)\right]^{\mathrm{2}} ={q}^{\mathrm{2}} {a}^{\mathrm{2}} \left(\mathrm{1}−{n}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\:{n}^{\mathrm{2}} \left\{\left({qb}−{R}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} {a}^{\mathrm{2}} \right\}+ \\ $$$$\:+\mathrm{2}\left({r}+\mathrm{5}\right)\left({qb}−{R}\right){n}+\left({r}+\mathrm{5}\right)^{\mathrm{2}} −{q}^{\mathrm{2}} {a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\mid{CM}\mid={R}\:=\mid{CS}\mid\:\:\Rightarrow \\ $$$$\:\left({ph}+\mathrm{3}{u}\right)^{\mathrm{2}} +\left({pk}−\mathrm{3}{v}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\: \\ $$$$\:\:\left[{ph}−\left({s}+\mathrm{5}\right){u}\right]^{\mathrm{2}} +\left[{pk}+\left({s}+\mathrm{5}\right){v}−{R}\right]^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\:\:{s}+\mathrm{5}\:\:{and}\:−\mathrm{3}\:\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\:\:\left({ph}−{ux}\right)^{\mathrm{2}} +\left({pk}+{vx}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:{sum}\:{of}\:{roots}\:\left({s}+\mathrm{5}\right)+\left(−\mathrm{3}\right)= \\ $$$$\:{s}+\mathrm{2}={phu}−{v}\left({pk}−{R}\right)\:\:\:\:\left({as}\:{h}^{\mathrm{2}} +{k}^{\mathrm{2}} =\mathrm{1}\right) \\ $$$$\&\:\:\:\:{u}^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……\left({v}\right),\left({vi}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${plus} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}\:,\:\:{h}^{\mathrm{2}} +{k}^{\mathrm{2}} =\mathrm{1}\:..\left({vii}\right)\&\left({viii}\right) \\ $$$$\:\:\mathrm{2}{q}={r}+\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({ix}\right) \\ $$$$\:\:\mathrm{4}{p}=\mathrm{3}\left({s}+\mathrm{5}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({x}\right) \\ $$$$\:\:\mathrm{5}\left({s}+\mathrm{3}\right)=\mathrm{6}\left({r}+\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({xi}\right) \\ $$$$\:\:{further}\:\:\:\:\mathrm{2}{bR}={q}+\mathrm{2}\:\:\:\&\:\:\:..\left({xii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{kR}={p}+\mathrm{4}\:\:\:\:\:\:….\left({xiii}\right) \\ $$$${Thirteen}\:{eqs}.\:{in}\:{thirteen}\:{unknowns} \\ $$$$\:{a},{b},{h},{k},{m},{n},{u},{v},{p},{q},{r},{s},\&{R}. \\ $$
Answered by mr W last updated on 30/Sep/19
Commented by mr W last updated on 30/Sep/19
$$\left({p}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({p}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\left({q}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}\left({R}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{b}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({q}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\left({r}+\mathrm{6}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left({R}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{c}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({r}+\mathrm{7}\right)^{\mathrm{2}} \\ $$$$\left({s}+\mathrm{5}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{4}\left({R}^{\mathrm{2}} −{d}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{d}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({s}+\mathrm{8}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${a}^{\mathrm{2}} +\left({p}−\frac{{p}+\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} +\left({q}−\frac{{q}+\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} +\left({p}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}{b}^{\mathrm{2}} +\left({q}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\left({p}+\mathrm{4}\right)^{\mathrm{2}} +\left({p}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({q}+\mathrm{2}\right)^{\mathrm{2}} +\left({q}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{p}={q} \\ $$$$ \\ $$$${b}^{\mathrm{2}} +\left(\frac{{q}+\mathrm{2}}{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} +\left(\frac{{r}+\mathrm{6}+\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{b}^{\mathrm{2}} +\left({q}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}{c}^{\mathrm{2}} +\left({r}+\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\left({q}+\mathrm{2}\right)^{\mathrm{2}} +\left({q}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({r}+\mathrm{7}\right)^{\mathrm{2}} +\left({r}+\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{q}={r}+\mathrm{6} \\ $$$$ \\ $$$${c}^{\mathrm{2}} +\left(\frac{{r}+\mathrm{6}+\mathrm{1}}{\mathrm{2}}−\mathrm{6}\right)^{\mathrm{2}} ={d}^{\mathrm{2}} +\left(\frac{{s}+\mathrm{5}+\mathrm{3}}{\mathrm{2}}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{c}^{\mathrm{2}} +\left({r}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{4}{d}^{\mathrm{2}} +\left({s}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\left({r}+\mathrm{7}\right)^{\mathrm{2}} +\left({r}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({s}+\mathrm{8}\right)^{\mathrm{2}} +\left({s}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{6}\left({r}+\mathrm{1}\right)=\mathrm{5}\left({s}+\mathrm{3}\right) \\ $$$$ \\ $$$${d}^{\mathrm{2}} +\left(\frac{{s}+\mathrm{5}+\mathrm{3}}{\mathrm{2}}−\mathrm{3}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\frac{{p}+\mathrm{4}}{\mathrm{2}}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{d}^{\mathrm{2}} +\left({s}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} +\left({p}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\left({s}+\mathrm{8}\right)^{\mathrm{2}} +\left({s}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({p}+\mathrm{4}\right)^{\mathrm{2}} +\left({p}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}\left({s}+\mathrm{5}\right)=\mathrm{4}{p} \\ $$$$ \\ $$$${q}=\mathrm{2}{p} \\ $$$${r}=\mathrm{2}{q}−\mathrm{6}=\mathrm{4}{p}−\mathrm{6} \\ $$$${s}=\frac{\mathrm{6}}{\mathrm{5}}\left({r}+\mathrm{1}\right)−\mathrm{3}=\frac{\mathrm{24}{p}}{\mathrm{5}}−\mathrm{9} \\ $$$$\mathrm{3}\left(\frac{\mathrm{24}{p}}{\mathrm{5}}−\mathrm{9}\right)=\mathrm{4}{p} \\ $$$$\mathrm{3}\left(\mathrm{24}{p}−\mathrm{45}\right)=\mathrm{20}{p} \\ $$$$\Rightarrow{p}=\frac{\mathrm{135}}{\mathrm{52}} \\ $$$$\Rightarrow{q}=\frac{\mathrm{135}}{\mathrm{26}} \\ $$$$\Rightarrow{r}=\mathrm{4}×\frac{\mathrm{135}}{\mathrm{52}}−\mathrm{6}=\frac{\mathrm{57}}{\mathrm{13}} \\ $$$$\Rightarrow{s}=\frac{\mathrm{24}}{\mathrm{5}}×\frac{\mathrm{135}}{\mathrm{52}}−\mathrm{9}=\frac{\mathrm{45}}{\mathrm{13}} \\ $$$$…… \\ $$
Commented by mr W last updated on 30/Sep/19
$${we}\:{can}\:{also}\:{get}\:{directly}: \\ $$$$\mathrm{5}\left({s}+\mathrm{3}\right)=\mathrm{6}\left({r}+\mathrm{1}\right) \\ $$$$\mathrm{4}{p}=\mathrm{3}\left({s}+\mathrm{5}\right) \\ $$$$\mathrm{2}{q}=\mathrm{1}×\left({r}+\mathrm{6}\right) \\ $$
Commented by ajfour last updated on 01/Oct/19
$${Thanks}\:{Sir},\:{can}\:{we}\:{now}\: \\ $$$${calculate}\:{Radius}\:? \\ $$
Commented by mr W last updated on 02/Oct/19
$${the}\:{radius}\:{should}\:{be}\:{unique},\:{but}\:{i} \\ $$$${have}\:{not}\:{got}\:{a}\:{way}\:{to}\:{calculate}. \\ $$