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Question Number 192966 by Mingma last updated on 01/Jun/23

Answered by aleks041103 last updated on 01/Jun/23

at the intersection (0,0)∈R^2 . the whole circle is (x−(5/4))^2 +(y+(1/2))^2 =((125)/(16))=(((5(√5))/4))^2 centers of A and B lie on the angle bisectors i.e. center of B lies on y=x=R ⇒(R−(5/4))^2 +(R+(1/2))^2 =(((5(√5))/4)−R)^2 2R^2 +((29)/(16))−(3/2)R=((125)/(16))+R^2 −((5(√5))/2)R R^2 +((5(√5)−3)/2)R−6=0 R=(1/2)(((3−5(√5))/2)±(√((((3−5(√5))/2))^2 +24))) =((3−5(√5))/4)+(1/4)(√(230−30(√5)))= =((3−5(√5))/4)+((15−(√5))/4)=((18−6(√5))/4)=((9−3(√5))/2) R_B =((9−3(√5))/2) for A the same. you can do the rest.

attheintersection(0,0)∈R2.thewholecircleis(x−54)2+(y+12)2=12516=(554)2centersofAandBlieontheanglebisectorsi.e.centerofBliesony=x=R⇒(R−54)2+(R+12)2=(554−R)22R2+2916−32R=12516+R2−552RR2+55−32R−6=0R=12(3−552±(3−552)2+24)=3−554+14230−305==3−554+15−54=18−654=9−352RB=9−352forAthesame.youcandotherest.

Commented by Mingma last updated on 01/Jun/23

Perfect ��

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