Question Number 224122 by fantastic last updated on 21/Aug/25
Commented by fantastic last updated on 21/Aug/25
$${Three}\:{circles}\:{in}\:{a}\:{big}\:{circle} \\ $$$${A}\:{pointed}\:{circle}\:{r}_{\mathrm{1}} \\ $$$${B}\:{pointed}\:{circle}\:{r}_{\mathrm{2}} \\ $$$${C}\:{pointed}\:{circle}\:{r}_{\mathrm{3}} \\ $$$${R}=? \\ $$$${Given}\::\measuredangle{ACB}=\mathrm{90}^{\mathrm{0}} \\ $$$$ \\ $$
Answered by fantastic last updated on 21/Aug/25
![AI=R−r_1 ,BI=R−r_2 ,CI=R−r_3 let ∡ACI=α cos α=((((r_1 +r_3 )^2 +(R−r_3 )^2 −(R−r_1 )^2 )/(2(r_1 +r_3 )(R−r_3 )))) cos (90^0 −α)=sin α=((((r_3 +r_2 )^2 +(R−r_3 )^2 −(R−r_2 )^2 )/(2(r_3 +r_2 )(R−r_3 )))) sin^2 α+cos^2 α=1 ⇒((((r_1 +r_3 )^2 +(R−r_3 )^2 −(R−r_1 )^2 )/(2(r_1 +r_3 )(R−r_3 ))))^2 +((((r_3 +r_2 )^2 +(R−r_3 )^2 −(R−r_2 )^2 )/(2(r_3 +r_2 )(R−r_3 ))))^2 =1 First part ⇒((((r_1 +r_3 )^2 +(R−r_3 )^2 −(R−r_1 )^2 )/(2(r_1 +r_3 )(R−r_3 ))))^2 =(((r_1 ^2 +r_3 ^2 +2r_1 r_3 +R^2 +r_3 ^2 −2Rr_3 −R^2 −r_1 ^2 +2Rr_1 )/(2(r_1 +r_3 )(R−r_3 ))))^2 =(((2(r_3 ^2 +r_1 r_3 −Rr_3 +Rr_1 ))/(2(r_1 +r_3 )(R−r_3 ))))^2 =(((r_3 (r_3 +r_1 )−R(r_3 −r_1 ))/((r_1 +r_3 )(R−r_3 ))))^2 =((r_3 /((R−r_3 )))−((R(r_3 −r_1 ))/((r_1 +r_3 )(R−r_3 ))))^2 Similarly ((((r_3 +r_2 )^2 +(R−r_3 )^2 −(R−r_2 )^2 )/(2(r_3 +r_2 )(R−r_3 ))))^2 =((r_2 /((R−r_3 )))−((R(r_3 −r_2 ))/((r_3 +r_2 )(R−r_3 ))))^2 let a=R−r_3 , b=r_3 −r_1 ,c=r_3 +r_1 , d=r_3 −r_2 ,e=r_3 +r_2 ((r_3 /a)−((Rb)/(ca)))^2 +((r_2 /a)−((Rd)/(ea)))^2 =1 ⇒(1/a^2 )(r_3 −((Rb)/c))^2 +(1/a^2 )(r_2 −((Rd)/e))^2 =1 ⇒(r_3 −((Rb)/c))^2 +(r_2 −((Rd)/e))^2 =a^2 ⇒r_3 ^2 +R^2 (b^2 /c^2 )−2R((r_3 b)/c)+r_2 ^2 +R^2 (d^2 /e^2 )−2R((r_2 d)/e)=a^2 ⇒R^2 ((b^2 /c^2 )+(d^2 /e^2 ))−2R(((r_3 b)/c)+((r_2 d)/e))=a^2 −(r_2 ^2 +r_3 ^2 ) ⇒R^2 ((b^2 /c^2 )+(d^2 /e^2 ))−2R(((r_3 b)/c)+((r_2 d)/e))=(R−r_3 )^2 −r_3 ^2 −r_2 ^2 ⇒R^2 ((b^2 /c^2 )+(d^2 /e^2 ))−2R(((r_3 b)/c)+((r_2 d)/e))=R^2 −2Rr_3 +r_3 ^2 −r_3 ^2 −r_2 ^2 ⇒R^2 ((b^2 /c^2 )+(d^2 /e^2 )−1)−2R(((r_3 b)/c)+((r_2 d)/e)−r_3 )=−r_2 ^2 ⇒R^2 ((b^2 /c^2 )+(d^2 /e^2 )−1)−R×2(((r_3 b)/c)+((r_2 d)/e)−r_3 )+r_2 ^2 =0 R=(((−(−2(((r_3 b)/c)+((r_2 d)/e)−r_3 ))+(√((−2(((r_3 b)/c)+((r_2 d)/e)−r_3 ))^2 −4((b^2 /c^2 )+(d^2 /e^2 )−1)r_2 ^2 )))/(2((b^2 /c^2 )+(d^2 /e^2 )−1))))[ negative one rejected] So R=(((2(((r_3 (r_3 −r_1 ))/((r_3 +r_1 )))+((r_2 (r_3 −r_2 ))/((r_3 +r_2 )))−r_3 )+(√(4(((r_3 (r_3 −r_1 ))/((r_3 +r_1 )))+((r_2 (r_3 −r_2 ))/((r_3 +r_2 )))−r_3 )^2 −4r_2 ^2 ((((r_3 −r_1 )^2 )/((r_3 +r_1 )^2 ))+(((r_3 −r_2 )^2 )/((r_3 +r_2 )^2 ))−1))))/(2((((r_3 −r_1 )^2 )/((r_3 +r_1 )^2 ))+(((r_3 −r_2 )^2 )/((r_3 +r_2 )^2 ))−1))))](https://www.tinkutara.com/question/Q224126.png)
$${AI}={R}−{r}_{\mathrm{1}} ,{BI}={R}−{r}_{\mathrm{2}} ,{CI}={R}−{r}_{\mathrm{3}} \\ $$$${let}\:\measuredangle{ACI}=\alpha \\ $$$$\mathrm{cos}\:\alpha=\left(\frac{\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({R}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)\: \\ $$$$\mathrm{cos}\:\left(\mathrm{90}^{\mathrm{0}} −\alpha\right)=\mathrm{sin}\:\alpha=\left(\frac{\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right) \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{cos}\:^{\mathrm{2}} \alpha=\mathrm{1} \\ $$$$\Rightarrow\left(\frac{\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({R}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} +\left(\frac{\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${First}\:{part} \\ $$$$\Rightarrow\left(\frac{\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({R}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{2}{r}_{\mathrm{1}} {r}_{\mathrm{3}} +{R}^{\mathrm{2}} +{r}_{\mathrm{3}} ^{\mathrm{2}} −\mathrm{2}{Rr}_{\mathrm{3}} −{R}^{\mathrm{2}} −{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{Rr}_{\mathrm{1}} }{\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{2}\left({r}_{\mathrm{3}} ^{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{3}} −{Rr}_{\mathrm{3}} +{Rr}_{\mathrm{1}} \right)}{\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} \: \\ $$$$=\left(\frac{{r}_{\mathrm{3}} \left({r}_{\mathrm{3}} +{r}_{\mathrm{1}} \right)−{R}\left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)}{\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{r}_{\mathrm{3}} }{\left({R}−{r}_{\mathrm{3}} \right)}−\frac{{R}\left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)}{\left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} \\ $$$${Similarly} \\ $$$$\left(\frac{\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{r}_{\mathrm{2}} }{\left({R}−{r}_{\mathrm{3}} \right)}−\frac{{R}\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)}{\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)\left({R}−{r}_{\mathrm{3}} \right)}\right)^{\mathrm{2}} \\ $$$${let}\:{a}={R}−{r}_{\mathrm{3}} \:,\:{b}={r}_{\mathrm{3}} −{r}_{\mathrm{1}} \:,{c}={r}_{\mathrm{3}} +{r}_{\mathrm{1}} ,\:{d}={r}_{\mathrm{3}} −{r}_{\mathrm{2}} ,{e}={r}_{\mathrm{3}} +{r}_{\mathrm{2}} \\ $$$$\left(\frac{{r}_{\mathrm{3}} }{{a}}−\frac{{Rb}}{{ca}}\right)^{\mathrm{2}} +\left(\frac{{r}_{\mathrm{2}} }{{a}}−\frac{{Rd}}{{ea}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\left({r}_{\mathrm{3}} −\frac{{Rb}}{{c}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\left({r}_{\mathrm{2}} −\frac{{Rd}}{{e}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\left({r}_{\mathrm{3}} −\frac{{Rb}}{{c}}\right)^{\mathrm{2}} +\left({r}_{\mathrm{2}} −\frac{{Rd}}{{e}}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\Rightarrow{r}_{\mathrm{3}} ^{\mathrm{2}} +{R}^{\mathrm{2}} \frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\mathrm{2}{R}\frac{{r}_{\mathrm{3}} {b}}{{c}}+{r}_{\mathrm{2}} ^{\mathrm{2}} +{R}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} }{{e}^{\mathrm{2}} }−\mathrm{2}{R}\frac{{r}_{\mathrm{2}} {d}}{{e}}={a}^{\mathrm{2}} \\ $$$$\Rightarrow{R}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} }{{e}^{\mathrm{2}} }\right)−\mathrm{2}{R}\left(\frac{{r}_{\mathrm{3}} {b}}{{c}}+\frac{{r}_{\mathrm{2}} {d}}{{e}}\right)={a}^{\mathrm{2}} −\left({r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{3}} ^{\mathrm{2}} \right) \\ $$$$\Rightarrow{R}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} }{{e}^{\mathrm{2}} }\right)−\mathrm{2}{R}\left(\frac{{r}_{\mathrm{3}} {b}}{{c}}+\frac{{r}_{\mathrm{2}} {d}}{{e}}\right)=\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −{r}_{\mathrm{3}} ^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow{R}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} }{{e}^{\mathrm{2}} }\right)−\mathrm{2}{R}\left(\frac{{r}_{\mathrm{3}} {b}}{{c}}+\frac{{r}_{\mathrm{2}} {d}}{{e}}\right)={R}^{\mathrm{2}} −\mathrm{2}{Rr}_{\mathrm{3}} +{r}_{\mathrm{3}} ^{\mathrm{2}} −{r}_{\mathrm{3}} ^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow{R}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} }{{e}^{\mathrm{2}} }−\mathrm{1}\right)−\mathrm{2}{R}\left(\frac{{r}_{\mathrm{3}} {b}}{{c}}+\frac{{r}_{\mathrm{2}} {d}}{{e}}−{r}_{\mathrm{3}} \right)=−{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow{R}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} }{{e}^{\mathrm{2}} }−\mathrm{1}\right)−{R}×\mathrm{2}\left(\frac{{r}_{\mathrm{3}} {b}}{{c}}+\frac{{r}_{\mathrm{2}} {d}}{{e}}−{r}_{\mathrm{3}} \right)+{r}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{0} \\ $$$${R}=\left(\frac{−\left(−\mathrm{2}\left(\frac{{r}_{\mathrm{3}} {b}}{{c}}+\frac{{r}_{\mathrm{2}} {d}}{{e}}−{r}_{\mathrm{3}} \right)\right)+\sqrt{\left(−\mathrm{2}\left(\frac{{r}_{\mathrm{3}} {b}}{{c}}+\frac{{r}_{\mathrm{2}} {d}}{{e}}−{r}_{\mathrm{3}} \right)\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} }{{e}^{\mathrm{2}} }−\mathrm{1}\right){r}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{2}\left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} }{{e}^{\mathrm{2}} }−\mathrm{1}\right)}\right)\left[\:{negative}\:{one}\:{rejected}\right] \\ $$$${So} \\ $$$${R}=\left(\frac{\mathrm{2}\left(\frac{{r}_{\mathrm{3}} \left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)}{\left({r}_{\mathrm{3}} +{r}_{\mathrm{1}} \right)}+\frac{{r}_{\mathrm{2}} \left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)}{\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)}−{r}_{\mathrm{3}} \right)+\sqrt{\mathrm{4}\left(\frac{{r}_{\mathrm{3}} \left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)}{\left({r}_{\mathrm{3}} +{r}_{\mathrm{1}} \right)}+\frac{{r}_{\mathrm{2}} \left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)}{\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{4}{r}_{\mathrm{2}} ^{\mathrm{2}} \left(\frac{\left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)^{\mathrm{2}} }{\left({r}_{\mathrm{3}} +{r}_{\mathrm{1}} \right)^{\mathrm{2}} }+\frac{\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} }−\mathrm{1}\right)}}{\mathrm{2}\left(\frac{\left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)^{\mathrm{2}} }{\left({r}_{\mathrm{3}} +{r}_{\mathrm{1}} \right)^{\mathrm{2}} }+\frac{\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} }−\mathrm{1}\right)}\right) \\ $$
Commented by mr W last updated on 21/Aug/25
$${in}\:{this}\:{case}\:{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} \:{and}\:{r}_{\mathrm{3}} \:{are}\: \\ $$$${dependent}\:{from}\:{each}\:{other}.\:{that} \\ $$$${means}\:{you}\:{only}\:{need}\:{to}\:{know}\:{r}_{\mathrm{1}} \:{and} \\ $$$${r}_{\mathrm{2}} . \\ $$$$\left({R}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right){R}−{r}_{\mathrm{1}} {r}_{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{R}=\frac{{r}_{\mathrm{1}} +{r}_{\mathrm{2}} +\sqrt{{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{6}{r}_{\mathrm{1}} {r}_{\mathrm{2}} }}{\mathrm{2}} \\ $$$${example}:\: \\ $$$${r}_{\mathrm{1}} =\mathrm{3},\:{r}_{\mathrm{2}} =\mathrm{2} \\ $$$${R}=\frac{\mathrm{3}+\mathrm{2}+\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{6}×\mathrm{3}×\mathrm{2}}}{\mathrm{2}}=\mathrm{6} \\ $$
Commented by mr W last updated on 21/Aug/25
$${you}\:{can}\:{get}\:{r}_{\mathrm{3}} \\ $$$${r}_{\mathrm{3}} =\frac{−{r}_{\mathrm{1}} −{r}_{\mathrm{2}} +\sqrt{{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{6}{r}_{\mathrm{1}} {r}_{\mathrm{2}} }}{\mathrm{2}}=\mathrm{1} \\ $$
Commented by mr W last updated on 22/Aug/25
$${generally} \\ $$$${R}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{1}}{{r}_{\mathrm{1}} {r}_{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{\mathrm{2}} {r}_{\mathrm{3}} }+\frac{\mathrm{1}}{{r}_{\mathrm{3}} {r}_{\mathrm{1}} }}−\frac{\mathrm{1}}{{r}_{\mathrm{1}} }−\frac{\mathrm{1}}{{r}_{\mathrm{2}} }−\frac{\mathrm{1}}{{r}_{\mathrm{3}} }} \\ $$$${you}\:{can}\:{get}\:{this}\:{by}\:{applying} \\ $$$${Descartes}'\:{theorem}. \\ $$
Commented by fantastic last updated on 21/Aug/25
$${ye} \\ $$
Commented by fantastic last updated on 21/Aug/25
$${is}\:{there}\:{a}\:{way}\:{to}\:{get}\:{this}\:{formula}. \\ $$
Commented by mr W last updated on 21/Aug/25
$${quite}\:{easy}.\:{see}\:{Q}\mathrm{77681}. \\ $$
Commented by fantastic last updated on 22/Aug/25
$${that}\:{is}\:{a}\:{great}\:{method}. \\ $$$${Sir}\:{how}\:{do}\:{you}\:{find}\:{these}\:{old} \\ $$$${questions}.{It}\:{will}\:{take}\:{a}\:{lot}\:{of} \\ $$$${time}\:{to}\:{do}\:{manually}. \\ $$
Commented by fantastic last updated on 22/Aug/25
$${Sir}\:{in}\:{Q}\mathrm{224082} \\ $$$${it}\:{says}\::{prove}\:{correctly}\:{that}\:{in}\: \\ $$$${coordinate}\:{plane}\left({XY}−{plane}\right) \\ $$$${you}\:{will}\:{not}\:{find}\:{a}\:{point}\:{in}\:{the}\:\mathrm{3}^{{rd}} \\ $$$${quadrant}\:{which}\:{has}\:{same}\:{distance} \\ $$$${from}\:{point}\:\left(\mathrm{5},\mathrm{4}\right)\:{and}\:\left(−\mathrm{2},\mathrm{3}\right). \\ $$$$ \\ $$$${Due}\:{to}\:{lack}\:{of}\:{time}\:{I}\:{wrote}\:{that} \\ $$$${the}\:\bot\:{bisector}\:{of}\:{the}\:{line}\:{connecting} \\ $$$${the}\:{two}\:{points}\:{will}\:{not}\:{go}\:{through}\:\mathrm{3}^{{rd}\:} {quadrant}. \\ $$$${am}\:{i}\:{right}\:{sir}. \\ $$
Commented by mr W last updated on 22/Aug/25
$${but}\:{it}\:{is}\:{not}\:{proved}\:{that}\:{this}\:{line} \\ $$$${doesn}'{t}\:{go}\:{through}\:{the}\:{third}\:{quadrant}. \\ $$
Commented by mr W last updated on 22/Aug/25
$${my}\:{solution}\:{were}: \\ $$$${assume}\:{there}\:{exists}\:{a}\:{point}\:\left({p},\:{q}\right) \\ $$$${in}\:{the}\:{third}\:{quadrant}\:{which}\:{has}\:{the} \\ $$$${same}\:{distance}\:{to}\:{the}\:{two}\:{points}, \\ $$$${then} \\ $$$$\sqrt{\left({p}−\mathrm{5}\right)^{\mathrm{2}} +\left({q}−\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\left({p}+\mathrm{2}\right)^{\mathrm{2}} +\left({q}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\left({p}−\mathrm{5}\right)^{\mathrm{2}} +\left({q}−\mathrm{4}\right)^{\mathrm{2}} =\left({p}+\mathrm{2}\right)^{\mathrm{2}} +\left({q}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\mathrm{7}{p}+{q}=\mathrm{14} \\ $$$${since}\:\left({p},\:{q}\right)\:{is}\:{in}\:{third}\:{quadrant},\: \\ $$$${p}\leqslant\mathrm{0},\:{q}\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{14}=\mathrm{7}{p}+{q}\leqslant\mathrm{0} \\ $$$${this}\:{is}\:{contradiction}.\: \\ $$
Commented by fantastic last updated on 22/Aug/25
$${hmm}… \\ $$$${you}\:{said}\:{right} \\ $$
Commented by fantastic last updated on 23/Aug/25
$${there}\:{is}.\:{but}\:{it}\:{matches}\:{with}\:{your}\:{solution} \\ $$
Commented by fantastic last updated on 23/Aug/25
Commented by fantastic last updated on 23/Aug/25
![Midpoint =(((5−2)/2) , ((4+3)/2))=((3/2),(7/2)) (((3−4)/(−2−5)))=(1/7) stiffness of the perpendicular line=−7 Y−(7/2)=−7(X−(3/2)) Y=14−7X...[eq.of ⊥ line] −a=14−7(−b) [(−b,−a) a point in 3rd quad.] −a=14+7b×× not possible](https://www.tinkutara.com/question/Q224163.png)
$${Midpoint}\:=\left(\frac{\mathrm{5}−\mathrm{2}}{\mathrm{2}}\:,\:\frac{\mathrm{4}+\mathrm{3}}{\mathrm{2}}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right) \\ $$$$\left(\frac{\mathrm{3}−\mathrm{4}}{−\mathrm{2}−\mathrm{5}}\right)=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$${stiffness}\:{of}\:{the}\:{perpendicular}\:{line}=−\mathrm{7} \\ $$$${Y}−\frac{\mathrm{7}}{\mathrm{2}}=−\mathrm{7}\left({X}−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${Y}=\mathrm{14}−\mathrm{7}{X}…\left[{eq}.{of}\:\bot\:{line}\right] \\ $$$$−{a}=\mathrm{14}−\mathrm{7}\left(−{b}\right)\:\:\:\left[\left(−{b},−{a}\right)\:\:\:\:{a}\:{point}\:{in}\:\mathrm{3}{rd}\:{quad}.\right] \\ $$$$−{a}=\mathrm{14}+\mathrm{7}{b}×× \\ $$$${not}\:{possible} \\ $$