Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 69681 by ajfour last updated on 26/Sep/19

Commented by ajfour last updated on 26/Sep/19

Q.69496   (A try..)

$${Q}.\mathrm{69496}\:\:\:\left({A}\:{try}..\right) \\ $$

Commented by ajfour last updated on 26/Sep/19

Radius of circle.

$${Radius}\:{of}\:{circle}. \\ $$

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.

$${let}\:\mathrm{cos}\:\alpha={a},\:\mathrm{cos}\:\beta={h},\:\mathrm{cos}\:\gamma={m}, \\ $$$$\mathrm{cos}\:\delta={u}\:\:\:\&\:\:\mathrm{sin}\:\alpha={b},\:\mathrm{sin}\:\beta={k}, \\ $$$$\mathrm{sin}\:\gamma={n},\:\mathrm{sin}\:\delta={v} \\ $$$${x}_{{D}} =−{qa}+{rm}={ph}−{su} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{su}}+\boldsymbol{{rm}}\:=\:\boldsymbol{{qa}}+\boldsymbol{{ph}}\:\:\:\:\:...\left({i}\right) \\ $$$${y}_{{D}} ={qb}+{rn}={pk}+{sv} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{sv}}−\boldsymbol{{rn}}\:=\:\boldsymbol{{qb}}−\boldsymbol{{pk}}\:\:\:\:\:\:...\left({ii}\right) \\ $$$$\mid{CP}\:\mid={R}\:=\mid{CT}\mid\:\Rightarrow \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\left(−{qa}−{m}\right)^{\mathrm{2}} +\left({qb}−{n}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\:\:\left[−{qa}+\left({r}+\mathrm{6}\right){m}\right]^{\mathrm{2}} +\left({qb}+\left({r}+\mathrm{6}\right){n}−{R}\right]^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\:{r}+\mathrm{6}\:{and}\:−\mathrm{1}\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\left(−{qa}+{mx}\right)^{\mathrm{2}} +\left({qb}+{nx}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$$$\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}{x}\left[−{qam}+{n}\left({qb}−{R}\right)\right] \\ $$$$\:+{q}^{\mathrm{2}} {a}^{\mathrm{2}} +{q}^{\mathrm{2}} {b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{qbR}−{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +\mathrm{2}{x}\left[−{qam}+{n}\left({qb}−{R}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{q}^{\mathrm{2}} −\mathrm{2}{qbR}=\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\:\:{r}+\mathrm{5}={qam}−{n}\left({qb}−{R}\right)\:\: \\ $$$$\&\:\:\:\:\:\:\:\:{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{1}\:\:\:\:\:\:\:\:.....\left({iii}\right),\left({iv}\right) \\ $$$$\:\left[{r}+\mathrm{5}+{n}\left({qb}−{R}\right)\right]^{\mathrm{2}} ={q}^{\mathrm{2}} {a}^{\mathrm{2}} \left(\mathrm{1}−{n}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\:{n}^{\mathrm{2}} \left\{\left({qb}−{R}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} {a}^{\mathrm{2}} \right\}+ \\ $$$$\:+\mathrm{2}\left({r}+\mathrm{5}\right)\left({qb}−{R}\right){n}+\left({r}+\mathrm{5}\right)^{\mathrm{2}} −{q}^{\mathrm{2}} {a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\mid{CM}\mid={R}\:=\mid{CS}\mid\:\:\Rightarrow \\ $$$$\:\left({ph}+\mathrm{3}{u}\right)^{\mathrm{2}} +\left({pk}−\mathrm{3}{v}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\: \\ $$$$\:\:\left[{ph}−\left({s}+\mathrm{5}\right){u}\right]^{\mathrm{2}} +\left[{pk}+\left({s}+\mathrm{5}\right){v}−{R}\right]^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\:\:{s}+\mathrm{5}\:\:{and}\:−\mathrm{3}\:\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\:\:\left({ph}−{ux}\right)^{\mathrm{2}} +\left({pk}+{vx}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:{sum}\:{of}\:{roots}\:\left({s}+\mathrm{5}\right)+\left(−\mathrm{3}\right)= \\ $$$$\:{s}+\mathrm{2}={phu}−{v}\left({pk}−{R}\right)\:\:\:\:\left({as}\:{h}^{\mathrm{2}} +{k}^{\mathrm{2}} =\mathrm{1}\right) \\ $$$$\&\:\:\:\:{u}^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......\left({v}\right),\left({vi}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${plus} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}\:,\:\:{h}^{\mathrm{2}} +{k}^{\mathrm{2}} =\mathrm{1}\:..\left({vii}\right)\&\left({viii}\right) \\ $$$$\:\:\mathrm{2}{q}={r}+\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({ix}\right) \\ $$$$\:\:\mathrm{4}{p}=\mathrm{3}\left({s}+\mathrm{5}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({x}\right) \\ $$$$\:\:\mathrm{5}\left({s}+\mathrm{3}\right)=\mathrm{6}\left({r}+\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({xi}\right) \\ $$$$\:\:{further}\:\:\:\:\mathrm{2}{bR}={q}+\mathrm{2}\:\:\:\&\:\:\:..\left({xii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{kR}={p}+\mathrm{4}\:\:\:\:\:\:....\left({xiii}\right) \\ $$$${Thirteen}\:{eqs}.\:{in}\:{thirteen}\:{unknowns} \\ $$$$\:{a},{b},{h},{k},{m},{n},{u},{v},{p},{q},{r},{s},\&{R}. \\ $$

Answered by mr W last updated on 30/Sep/19

Commented by mr W last updated on 30/Sep/19

(p+4)^2 =4(R^2 −a^2 )  ⇒4a^2 =4R^2 −(p+4)^2   (q+2)^2 =4(R^2 −b^2 )  ⇒4b^2 =4R^2 −(q+2)^2   (r+6+1)^2 =4(R^2 −c^2 )  ⇒4c^2 =4R^2 −(r+7)^2   (s+5+3)^2 =4(R^2 −d^2 )  ⇒4d^2 =4R^2 −(s+8)^2     a^2 +(p−((p+4)/2))^2 =b^2 +(q−((q+2)/2))^2   4a^2 +(p−4)^2 =4b^2 +(q−2)^2   4R^2 −(p+4)^2 +(p−4)^2 =4R^2 −(q+2)^2 +(q−2)^2   ⇒2p=q    b^2 +(((q+2)/2)−2)^2 =c^2 +(((r+6+1)/2)−1)^2   4b^2 +(q−2)^2 =4c^2 +(r+5)^2   4R^2 −(q+2)^2 +(q−2)^2 =4R^2 −(r+7)^2 +(r+5)^2   ⇒2q=r+6    c^2 +(((r+6+1)/2)−6)^2 =d^2 +(((s+5+3)/2)−5)^2   4c^2 +(r−5)^2 =4d^2 +(s−2)^2   4R^2 −(r+7)^2 +(r−5)^2 =4R^2 −(s+8)^2 +(s−2)^2   ⇒6(r+1)=5(s+3)    d^2 +(((s+5+3)/2)−3)^2 =a^2 +(((p+4)/2)−4)^2   4d^2 +(s+2)^2 =4a^2 +(p−4)^2   4R^2 −(s+8)^2 +(s+2)^2 =4R^2 −(p+4)^2 +(p−4)^2   ⇒3(s+5)=4p    q=2p  r=2q−6=4p−6  s=(6/5)(r+1)−3=((24p)/5)−9  3(((24p)/5)−9)=4p  3(24p−45)=20p  ⇒p=((135)/(52))  ⇒q=((135)/(26))  ⇒r=4×((135)/(52))−6=((57)/(13))  ⇒s=((24)/5)×((135)/(52))−9=((45)/(13))  ......

$$\left({p}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({p}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\left({q}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}\left({R}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{b}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({q}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\left({r}+\mathrm{6}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left({R}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{c}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({r}+\mathrm{7}\right)^{\mathrm{2}} \\ $$$$\left({s}+\mathrm{5}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{4}\left({R}^{\mathrm{2}} −{d}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{d}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({s}+\mathrm{8}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${a}^{\mathrm{2}} +\left({p}−\frac{{p}+\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} +\left({q}−\frac{{q}+\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} +\left({p}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}{b}^{\mathrm{2}} +\left({q}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\left({p}+\mathrm{4}\right)^{\mathrm{2}} +\left({p}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({q}+\mathrm{2}\right)^{\mathrm{2}} +\left({q}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{p}={q} \\ $$$$ \\ $$$${b}^{\mathrm{2}} +\left(\frac{{q}+\mathrm{2}}{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} +\left(\frac{{r}+\mathrm{6}+\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{b}^{\mathrm{2}} +\left({q}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}{c}^{\mathrm{2}} +\left({r}+\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\left({q}+\mathrm{2}\right)^{\mathrm{2}} +\left({q}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({r}+\mathrm{7}\right)^{\mathrm{2}} +\left({r}+\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{q}={r}+\mathrm{6} \\ $$$$ \\ $$$${c}^{\mathrm{2}} +\left(\frac{{r}+\mathrm{6}+\mathrm{1}}{\mathrm{2}}−\mathrm{6}\right)^{\mathrm{2}} ={d}^{\mathrm{2}} +\left(\frac{{s}+\mathrm{5}+\mathrm{3}}{\mathrm{2}}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{c}^{\mathrm{2}} +\left({r}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{4}{d}^{\mathrm{2}} +\left({s}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\left({r}+\mathrm{7}\right)^{\mathrm{2}} +\left({r}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({s}+\mathrm{8}\right)^{\mathrm{2}} +\left({s}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{6}\left({r}+\mathrm{1}\right)=\mathrm{5}\left({s}+\mathrm{3}\right) \\ $$$$ \\ $$$${d}^{\mathrm{2}} +\left(\frac{{s}+\mathrm{5}+\mathrm{3}}{\mathrm{2}}−\mathrm{3}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\frac{{p}+\mathrm{4}}{\mathrm{2}}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{d}^{\mathrm{2}} +\left({s}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} +\left({p}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\left({s}+\mathrm{8}\right)^{\mathrm{2}} +\left({s}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} −\left({p}+\mathrm{4}\right)^{\mathrm{2}} +\left({p}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}\left({s}+\mathrm{5}\right)=\mathrm{4}{p} \\ $$$$ \\ $$$${q}=\mathrm{2}{p} \\ $$$${r}=\mathrm{2}{q}−\mathrm{6}=\mathrm{4}{p}−\mathrm{6} \\ $$$${s}=\frac{\mathrm{6}}{\mathrm{5}}\left({r}+\mathrm{1}\right)−\mathrm{3}=\frac{\mathrm{24}{p}}{\mathrm{5}}−\mathrm{9} \\ $$$$\mathrm{3}\left(\frac{\mathrm{24}{p}}{\mathrm{5}}−\mathrm{9}\right)=\mathrm{4}{p} \\ $$$$\mathrm{3}\left(\mathrm{24}{p}−\mathrm{45}\right)=\mathrm{20}{p} \\ $$$$\Rightarrow{p}=\frac{\mathrm{135}}{\mathrm{52}} \\ $$$$\Rightarrow{q}=\frac{\mathrm{135}}{\mathrm{26}} \\ $$$$\Rightarrow{r}=\mathrm{4}×\frac{\mathrm{135}}{\mathrm{52}}−\mathrm{6}=\frac{\mathrm{57}}{\mathrm{13}} \\ $$$$\Rightarrow{s}=\frac{\mathrm{24}}{\mathrm{5}}×\frac{\mathrm{135}}{\mathrm{52}}−\mathrm{9}=\frac{\mathrm{45}}{\mathrm{13}} \\ $$$$...... \\ $$

Commented by mr W last updated on 30/Sep/19

we can also get directly:  5(s+3)=6(r+1)  4p=3(s+5)  2q=1×(r+6)

$${we}\:{can}\:{also}\:{get}\:{directly}: \\ $$$$\mathrm{5}\left({s}+\mathrm{3}\right)=\mathrm{6}\left({r}+\mathrm{1}\right) \\ $$$$\mathrm{4}{p}=\mathrm{3}\left({s}+\mathrm{5}\right) \\ $$$$\mathrm{2}{q}=\mathrm{1}×\left({r}+\mathrm{6}\right) \\ $$

Commented by ajfour last updated on 01/Oct/19

Thanks Sir, can we now   calculate Radius ?

$${Thanks}\:{Sir},\:{can}\:{we}\:{now}\: \\ $$$${calculate}\:{Radius}\:? \\ $$

Commented by mr W last updated on 02/Oct/19

the radius should be unique, but i  have not got a way to calculate.

$${the}\:{radius}\:{should}\:{be}\:{unique},\:{but}\:{i} \\ $$$${have}\:{not}\:{got}\:{a}\:{way}\:{to}\:{calculate}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com