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Question Number 77990 by peter frank last updated on 12/Jan/20
Find the equation to the two circles each of which touch the three circle x^2 +y^2 =4a^2 x^2 +y^2 +2ax=0 x^2 +y^2 −2ax=0
$${Find}\:{the}\:{equation}\:{to}\:{the} \\ $$$${two}\:{circles}\:{each}\:{of} \\ $$$${which}\:{touch}\:{the}\:{three}\:{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{ax}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{ax}=\mathrm{0} \\ $$$$ \\ $$
Answered by john santu last updated on 18/Jan/20
(1)x^2 +y^2 =(2a)^2 , C_1 (0,0) ,r_1 =2a (2)(x+a)^2 +y^2 =a^2 ,C_2 (−a,0) ,r_2 =a (3) x^2 +(y−a)^2 =a^2 , C_3 (0,a) ,r_3 =a The circle equation that will look for is centered at point (p,q) with radius r_4 ⇒r_4 =(√(p^2 +q^2 )) , r_4 = (√((p+a)^2 +q^2 )), r_4 =(√(p^2 +(q−a)^2 ))
$$\left(\mathrm{1}\right){x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{2}{a}\right)^{\mathrm{2}} \:,\:{C}_{\mathrm{1}} \left(\mathrm{0},\mathrm{0}\right)\:,{r}_{\mathrm{1}} =\mathrm{2}{a} \\ $$$$\left(\mathrm{2}\right)\left({x}+{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \:,{C}_{\mathrm{2}} \left(−{a},\mathrm{0}\right)\:,{r}_{\mathrm{2}} ={a} \\ $$$$\left(\mathrm{3}\right)\:{x}^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \:,\:{C}_{\mathrm{3}} \left(\mathrm{0},{a}\right)\:,{r}_{\mathrm{3}} ={a} \\ $$$${The}\:{circle}\:{equation}\:{that}\:{will}\: \\ $$$${look}\:{for}\:{is}\:{centered}\:{at}\:{point} \\ $$$$\left({p},{q}\right)\:{with}\:{radius}\:{r}_{\mathrm{4}} \\ $$$$\Rightarrow{r}_{\mathrm{4}} =\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} \:}\:,\:{r}_{\mathrm{4}} =\:\sqrt{\left({p}+{a}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} \:},\:{r}_{\mathrm{4}} =\sqrt{{p}^{\mathrm{2}} +\left({q}−{a}\right)^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by john santu last updated on 18/Jan/20
⇒p^2 +q^2 =p^2 +2ap+a^2 +q^2 p = −(a/2). ⇒p^2 +q^2 =p^2 +q^2 −2aq+a^2 q = (a/2) . now we calculate r_4 = (√((1/4)a^2 +(1/4)a^2 )) = (1/2)a(√2). finally ∴ (x+(a/2))^2 +(y−(a/2))^2 =(a^2 /2)
$$\Rightarrow{p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={p}^{\mathrm{2}} +\mathrm{2}{ap}+{a}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$${p}\:=\:−\frac{{a}}{\mathrm{2}}. \\ $$$$\Rightarrow{p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{aq}+{a}^{\mathrm{2}} \\ $$$${q}\:=\:\frac{{a}}{\mathrm{2}}\:.\:{now}\:{we}\:{calculate}\: \\ $$$${r}_{\mathrm{4}} \:=\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}{a}\sqrt{\mathrm{2}}. \\ $$$${finally}\:\therefore\:\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 18/Jan/20
sir: but (red) circle (x+(a/2))^2 +(y−(a/2))^2 =(a^2 /2) doesn′t touch the three given circles!
$${sir}: \\ $$$${but}\:\left({red}\right)\:{circle}\:\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${doesn}'{t}\:{touch}\:{the}\:{three}\:{given}\:{circles}! \\ $$
Commented by mr W last updated on 18/Jan/20
Commented by john santu last updated on 19/Jan/20
sorry mister, that instead i circle through the center of three circles given.
$${sorry}\:{mister},\:{that}\:{instead}\:{i}\:{circle} \\ $$$${through}\:{the}\:{center}\:{of}\:{three}\:{circles}\: \\ $$$${given}. \\ $$
Answered by mr W last updated on 18/Jan/20
x^2 +y^2 =(2a)^2 ⇒circle (0,0), radius 2a (x+a)^2 +y^2 =a^2 ⇒circle (−a,0), radius a (x−a)^2 +y^2 =a^2 ⇒circle (a,0), radius a fourth circle (0,c),radius r c+r=2a c^2 +a^2 =(a+r)^2 4a^2 −4ar+r^2 +a^2 =a^2 +r^2 +2ar 4a^2 =6ar ⇒r=((2a)/3) ⇒c=2a−((2a)/3)=((4a)/3) eqn. of fourth circle (two possibilties): x^2 +(y±((4a)/3))^2 =(((2a)/3))^2
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{2}{a}\right)^{\mathrm{2}} \:\Rightarrow{circle}\:\left(\mathrm{0},\mathrm{0}\right),\:{radius}\:\mathrm{2}{a} \\ $$$$\left({x}+{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \Rightarrow{circle}\:\left(−{a},\mathrm{0}\right),\:{radius}\:{a} \\ $$$$\left({x}−{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \Rightarrow{circle}\:\left({a},\mathrm{0}\right),\:{radius}\:{a} \\ $$$${fourth}\:{circle}\:\left(\mathrm{0},{c}\right),{radius}\:{r} \\ $$$${c}+{r}=\mathrm{2}{a} \\ $$$${c}^{\mathrm{2}} +{a}^{\mathrm{2}} =\left({a}+{r}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{ar}+{r}^{\mathrm{2}} +{a}^{\mathrm{2}} ={a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} =\mathrm{6}{ar} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\Rightarrow{c}=\mathrm{2}{a}−\frac{\mathrm{2}{a}}{\mathrm{3}}=\frac{\mathrm{4}{a}}{\mathrm{3}} \\ $$$${eqn}.\:{of}\:{fourth}\:{circle}\:\left({two}\:{possibilties}\right): \\ $$$${x}^{\mathrm{2}} +\left({y}\pm\frac{\mathrm{4}{a}}{\mathrm{3}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{2}{a}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$
Commented by mr W last updated on 18/Jan/20
Commented by peter frank last updated on 18/Jan/20
where c+r=2a ? come from sir
$${where}\:{c}+{r}=\mathrm{2}{a}\:?\:{come}\:{from}\:{sir} \\ $$
Commented by mr W last updated on 18/Jan/20
Commented by mr W last updated on 18/Jan/20
OC+CB=OB ⇒c+r=2a OC^2 +OA^2 =AC^2 ⇒c^2 +a^2 =(a+r)^2
$${OC}+{CB}={OB} \\ $$$$\Rightarrow{c}+{r}=\mathrm{2}{a} \\ $$$${OC}^{\mathrm{2}} +{OA}^{\mathrm{2}} ={AC}^{\mathrm{2}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} +{a}^{\mathrm{2}} =\left({a}+{r}\right)^{\mathrm{2}} \\ $$
Commented by john santu last updated on 19/Jan/20
good sir
$${good}\:{sir} \\ $$

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