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Question Number 37263 by ajfour last updated on 11/Jun/18

Commented by ajfour last updated on 11/Jun/18

If the circles (radii given) touch in the manner shown above, Find coordinates of centre C in terms of a, b, and R.

Ifthecircles(radiigiven)touchinthemannershownabove,FindcoordinatesofcentreCintermsofa,b,andR.

Answered by MrW3 last updated on 11/Jun/18

(x_C +a)^2 +y_C ^2 =(R−a)^2 ...(i) (x_C −b)^2 +y_C ^2 =(R−b)^2 ...(ii) (i)−(ii): (x_C +a)^2 −(x_C −b)^2 =(R−a)^2 −(R−b)^2 (a+b)(2x_C +a−b)=(2R−a−b)(−a+b) 2(a+b)x_C +(a+b)(a−b)=−2R(a−b)+(a+b)(a−b) 2(a+b)x_C =−2R(a−b) ⇒x_C =−(((a−b)R)/(a+b)) from (i): y_C ^2 =(R−a)^2 −(x_C +a)^2 =(R+x_C )(R−2a−x_C ) =[R−(((a−b)R)/(a+b))][R−2a+(((a−b)R)/(a+b))] =(((Ra+Rb−Ra+Rb)(Ra+Rb−2a^2 −2ab+Ra−Rb))/((a+b)^2 )) =((4Rab(R−a−b))/((a+b)^2 )) ⇒y_C =±((2(√(ab(R−a−b)R)))/(a+b))

(xC+a)2+yC2=(R−a)2...(i)(xC−b)2+yC2=(R−b)2...(ii)(i)−(ii):(xC+a)2−(xC−b)2=(R−a)2−(R−b)2(a+b)(2xC+a−b)=(2R−a−b)(−a+b)2(a+b)xC+(a+b)(a−b)=−2R(a−b)+(a+b)(a−b)2(a+b)xC=−2R(a−b)⇒xC=−(a−b)Ra+bfrom(i):yC2=(R−a)2−(xC+a)2=(R+xC)(R−2a−xC)=[R−(a−b)Ra+b][R−2a+(a−b)Ra+b]=(Ra+Rb−Ra+Rb)(Ra+Rb−2a2−2ab+Ra−Rb)(a+b)2=4Rab(R−a−b)(a+b)2⇒yC=±2ab(R−a−b)Ra+b

Commented by ajfour last updated on 11/Jun/18

Excellent again sir. y_C =0 if a+b=R . Fine enough.

Excellentagainsir.yC=0ifa+b=R.Fineenough.

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