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Question Number 54270 by peter frank last updated on 01/Feb/19

Find the equation to two circles which touch the x−axis at the origin and also touch the line 12x+5y=60

$${Find}\:{the}\:{equation}\:{to}\:{two} \\ $$$${circles}\:{which}\:{touch}\:{the}\: \\ $$$${x}−{axis}\:{at}\:{the}\:{origin} \\ $$$${and}\:{also}\:{touch}\:{the}\:{line} \\ $$$$\mathrm{12}{x}+\mathrm{5}{y}=\mathrm{60} \\ $$

Answered by ajfour last updated on 01/Feb/19

x^2 +(y−k)^2 =k^2 c=k±k (√(1+m^2 )) ⇒ 12=k±k(√(1+(((12)/5))^2 )) ⇒ k = ((12)/(1±((13)/5))) = ((10)/3), −((15)/2) hence eq. of circle above x-axis is 3x^2 +3y^2 −20y=0 while that below x-axis is x^2 +y^2 +15y=0 .

$${x}^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={k}^{\mathrm{2}} \\ $$$${c}={k}\pm{k}\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }\:\Rightarrow\:\mathrm{12}={k}\pm{k}\sqrt{\mathrm{1}+\left(\frac{\mathrm{12}}{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$$\:\Rightarrow\:\:{k}\:=\:\frac{\mathrm{12}}{\mathrm{1}\pm\frac{\mathrm{13}}{\mathrm{5}}}\:=\:\frac{\mathrm{10}}{\mathrm{3}},\:−\frac{\mathrm{15}}{\mathrm{2}} \\ $$$${hence}\:{eq}.\:{of}\:{circle}\:{above}\:{x}-{axis}\:{is} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{20}{y}=\mathrm{0} \\ $$$${while}\:{that}\:{below}\:{x}-{axis}\:{is} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{15}{y}=\mathrm{0}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19

x^2 +(y−a)^2 =a^2 centre(0,a) ∣((12×0+5a−60)/(√(12^2 +5^2 )))∣=a 5a−60=13a a=((−60)/8)=((−15)/2) x^2 +(y+7.5)^2 =(7.5)^2 edited for second equatiin centre(0,−a) x^2 +(y+a)^2 =a^2 ∣((12×0+5(−a)−60)/(√(12^2 +5^2 )))∣=a −5a−60=13a 18a=−60 a=((−10)/3) x^2 +(y−((10)/3))^2 =((100)/9) x^2 +y^2 −((20y)/3)=0 3x^2 +3y^2 −20y=0

$${x}^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \:{centre}\left(\mathrm{0},{a}\right) \\ $$$$\mid\frac{\mathrm{12}×\mathrm{0}+\mathrm{5}{a}−\mathrm{60}}{\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }}\mid={a} \\ $$$$\mathrm{5}{a}−\mathrm{60}=\mathrm{13}{a} \\ $$$${a}=\frac{−\mathrm{60}}{\mathrm{8}}=\frac{−\mathrm{15}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left({y}+\mathrm{7}.\mathrm{5}\right)^{\mathrm{2}} =\left(\mathrm{7}.\mathrm{5}\right)^{\mathrm{2}} \\ $$$${edited}\:{for}\:{second}\:{equatiin}\:{centre}\left(\mathrm{0},−{a}\right) \\ $$$${x}^{\mathrm{2}} +\left({y}+{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mid\frac{\mathrm{12}×\mathrm{0}+\mathrm{5}\left(−{a}\right)−\mathrm{60}}{\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }}\mid={a} \\ $$$$−\mathrm{5}{a}−\mathrm{60}=\mathrm{13}{a} \\ $$$$\mathrm{18}{a}=−\mathrm{60}\:\:{a}=\frac{−\mathrm{10}}{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} +\left({y}−\frac{\mathrm{10}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{100}}{\mathrm{9}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\frac{\mathrm{20}{y}}{\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{20}{y}=\mathrm{0} \\ $$

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