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Question-84409

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Question Number 84409 by mr W last updated on 12/Mar/20
Commented by mr W last updated on 12/Mar/20
Find the radius r.
$${Find}\:{the}\:{radius}\:{r}. \\ $$
Commented by ajfour last updated on 12/Mar/20
Commented by ajfour last updated on 12/Mar/20
x^2 +y^2 =(2+r)^2 (x−5)^2 +y^2 =(3+r)^2 5(5−2x)=5+2r x=2−(r/5) , y=(√((2+r)^2 −(2−(r/5))^2 )) A{((2(2−r/5))/((2+r))), ((2y)/(2+r))} B{x+((r(3−x))/(3+r)), ((3y)/(3+r))} C(2,0) cos α=((2^2 +2^2 −s^2 )/(2×2^2 ))=(((2+r)^2 +5^2 −(3+r)^2 )/(2×5(2+r))) .......(i) s^2 ={2−((2/(2+r)))(2−(r/5))}^2 +(4/((2+r)^2 )){(2+r)^2 −(2−(r/5))^2 } ........(ii) using (ii) in (i) we get r=1 .
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{2}+{r}\right)^{\mathrm{2}} \\ $$$$\left({x}−\mathrm{5}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{3}+{r}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}\left(\mathrm{5}−\mathrm{2}{x}\right)=\mathrm{5}+\mathrm{2}{r} \\ $$$${x}=\mathrm{2}−\frac{{r}}{\mathrm{5}}\:\:,\:\:\:{y}=\sqrt{\left(\mathrm{2}+{r}\right)^{\mathrm{2}} −\left(\mathrm{2}−\frac{{r}}{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$${A}\left\{\frac{\mathrm{2}\left(\mathrm{2}−{r}/\mathrm{5}\right)}{\left(\mathrm{2}+{r}\right)},\:\:\frac{\mathrm{2}{y}}{\mathrm{2}+{r}}\right\} \\ $$$${B}\left\{{x}+\frac{{r}\left(\mathrm{3}−{x}\right)}{\mathrm{3}+{r}},\:\frac{\mathrm{3}{y}}{\mathrm{3}+{r}}\right\} \\ $$$${C}\left(\mathrm{2},\mathrm{0}\right) \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −{s}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}^{\mathrm{2}} }=\frac{\left(\mathrm{2}+{r}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\left(\mathrm{3}+{r}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}\left(\mathrm{2}+{r}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….\left({i}\right) \\ $$$${s}^{\mathrm{2}} =\left\{\mathrm{2}−\left(\frac{\mathrm{2}}{\mathrm{2}+{r}}\right)\left(\mathrm{2}−\frac{{r}}{\mathrm{5}}\right)\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:+\frac{\mathrm{4}}{\left(\mathrm{2}+{r}\right)^{\mathrm{2}} }\left\{\left(\mathrm{2}+{r}\right)^{\mathrm{2}} −\left(\mathrm{2}−\frac{{r}}{\mathrm{5}}\right)^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:……..\left({ii}\right) \\ $$$${using}\:\left({ii}\right)\:{in}\:\left({i}\right) \\ $$$${we}\:{get}\:\:\boldsymbol{{r}}=\mathrm{1}\:. \\ $$
Commented by mr W last updated on 13/Mar/20
thanks for showing a different way sir!
$${thanks}\:{for}\:{showing}\:{a}\:{different}\:{way} \\ $$$${sir}! \\ $$
Answered by M±th+et£s last updated on 12/Mar/20
Answered by M±th+et£s last updated on 12/Mar/20
s=(1/2)p=(5+3+r+2+r)=r+5 A=(√(s(s−5)(s−r−3)(s−r−2)))=(√(6r(r+5))) r=(A/s)=((√(6r(r+5)))/(r+5))⇒r^2 =((6r)/(r+5))⇒r^2 +5r−6=0 r=1
$${s}=\frac{\mathrm{1}}{\mathrm{2}}{p}=\left(\mathrm{5}+\mathrm{3}+{r}+\mathrm{2}+{r}\right)={r}+\mathrm{5} \\ $$$${A}=\sqrt{{s}\left({s}−\mathrm{5}\right)\left({s}−{r}−\mathrm{3}\right)\left({s}−{r}−\mathrm{2}\right)}=\sqrt{\mathrm{6}{r}\left({r}+\mathrm{5}\right)} \\ $$$${r}=\frac{{A}}{{s}}=\frac{\sqrt{\mathrm{6}{r}\left({r}+\mathrm{5}\right)}}{{r}+\mathrm{5}}\Rightarrow{r}^{\mathrm{2}} =\frac{\mathrm{6}{r}}{{r}+\mathrm{5}}\Rightarrow{r}^{\mathrm{2}} +\mathrm{5}{r}−\mathrm{6}=\mathrm{0}\: \\ $$$${r}=\mathrm{1} \\ $$
Commented by mr W last updated on 12/Mar/20
thanks! you assumed that the green circle is the incircle, this should be proved at first.
$${thanks}! \\ $$$${you}\:{assumed}\:{that}\:{the}\:{green}\:{circle}\:{is} \\ $$$${the}\:{incircle},\:{this}\:{should}\:{be}\:{proved}\:{at} \\ $$$${first}. \\ $$
Answered by mr W last updated on 12/Mar/20
Commented by mr W last updated on 12/Mar/20
A,B,C,D are the centers of the circles. E,F,G are the touching points. It can be proved that E lies on AB, F on BC and G on CA. AG=AE=radius of circle A=a=2 DG=DE=radius of green circle D=r ⇒ΔAGD≡ΔAED ⇒∠GAD=∠EAD ⇒AD is bisector of angle ∠A. similarly ⇒BD is bisector of angle ∠B. ⇒CD is bisector of angle ∠C. ⇒circle D is incircle of triangle ABC.
$${A},{B},{C},{D}\:{are}\:{the}\:{centers}\:{of}\:{the}\:{circles}. \\ $$$${E},{F},{G}\:{are}\:{the}\:{touching}\:{points}. \\ $$$$ \\ $$$${It}\:{can}\:{be}\:{proved}\:{that}\:{E}\:{lies}\:{on}\:{AB}, \\ $$$${F}\:{on}\:{BC}\:{and}\:{G}\:{on}\:{CA}. \\ $$$$ \\ $$$${AG}={AE}={radius}\:{of}\:{circle}\:{A}={a}=\mathrm{2} \\ $$$${DG}={DE}={radius}\:{of}\:{green}\:{circle}\:{D}={r} \\ $$$$\Rightarrow\Delta{AGD}\equiv\Delta{AED} \\ $$$$\Rightarrow\angle{GAD}=\angle{EAD} \\ $$$$\Rightarrow{AD}\:{is}\:{bisector}\:{of}\:{angle}\:\angle{A}. \\ $$$${similarly} \\ $$$$\Rightarrow{BD}\:{is}\:{bisector}\:{of}\:{angle}\:\angle{B}. \\ $$$$\Rightarrow{CD}\:{is}\:{bisector}\:{of}\:{angle}\:\angle{C}. \\ $$$$ \\ $$$$\Rightarrow{circle}\:{D}\:{is}\:{incircle}\:{of}\:{triangle}\:{ABC}. \\ $$
Commented by mr W last updated on 12/Mar/20
a=radius of circle A=2 b=radius of circle B=3 c=radius of circle C=r d=radius of circle D=r s=a+b+c Δ_(ABC) =(√(s(s−a−b)(s−b−c)(s−c−a))) Δ_(ABC) =(√((a+b+c)abc)) d=(Δ_(ABC) /s) ⇒d=(√((abc)/(a+b+c))) in our case: r=(√((2×3×r)/(2+3+r))) r(5+r)=6 ⇒(r+6)(r−1)=0 ⇒r=1
$${a}={radius}\:{of}\:{circle}\:{A}=\mathrm{2} \\ $$$${b}={radius}\:{of}\:{circle}\:{B}=\mathrm{3} \\ $$$${c}={radius}\:{of}\:{circle}\:{C}={r} \\ $$$${d}={radius}\:{of}\:{circle}\:{D}={r} \\ $$$${s}={a}+{b}+{c} \\ $$$$\Delta_{{ABC}} =\sqrt{{s}\left({s}−{a}−{b}\right)\left({s}−{b}−{c}\right)\left({s}−{c}−{a}\right)} \\ $$$$\Delta_{{ABC}} =\sqrt{\left({a}+{b}+{c}\right){abc}} \\ $$$${d}=\frac{\Delta_{{ABC}} }{{s}} \\ $$$$\Rightarrow{d}=\sqrt{\frac{{abc}}{{a}+{b}+{c}}} \\ $$$$ \\ $$$${in}\:{our}\:{case}: \\ $$$${r}=\sqrt{\frac{\mathrm{2}×\mathrm{3}×{r}}{\mathrm{2}+\mathrm{3}+{r}}} \\ $$$${r}\left(\mathrm{5}+{r}\right)=\mathrm{6}\:\Rightarrow\left({r}+\mathrm{6}\right)\left({r}−\mathrm{1}\right)=\mathrm{0}\:\Rightarrow{r}=\mathrm{1} \\ $$
Commented by ajfour last updated on 12/Mar/20
but sir, i should get the same answer, as per my working..
$${but}\:{sir},\:{i}\:{should}\:{get}\:{the}\:{same} \\ $$$${answer},\:{as}\:{per}\:{my}\:{working}.. \\ $$
Commented by mr W last updated on 12/Mar/20
yes. you should get the same answer.
$${yes}.\:{you}\:{should}\:{get}\:{the}\:{same}\:{answer}. \\ $$

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