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Question-159392




Question Number 159392 by ajfour last updated on 16/Nov/21
Commented by ajfour last updated on 16/Nov/21
  Find radius of the circles(equal).
Findradiusofthecircles(equal).
Answered by mr W last updated on 17/Nov/21
Commented by mr W last updated on 18/Nov/21
eqn. of OC:  say y=mx ⇒mx−y=0  eqn. of AD:  sau y=−k(x−1) ⇒kx+y−k=0  P(r,1−r)  Q(q,r)    r=−((mr−1+r)/( (√(1+m^2 ))))   ...(i)  r=((mq−r)/( (√(1+m^2 ))))   ...(ii)  r=−((kr+1−r−k)/( (√(1+k^2 ))))   ...(iii)  r=−((kq+r−k)/( (√(1+k^2 ))))   ...(iv)  4 unknowns: r, m, k, q    mq−r+mr−1+r=0  ⇒q=(1/m)−r  kq+r−k=kr+1−r−k  ⇒k=((1−2r)/(q−r))  k=((1−2r)/((1/m)−2r))   ... (I)    from (ii):  r=((1−(m+1)r)/( (√(1+m^2 ))))  r=(1/(1+m+(√(1+m^2 ))))   ...(II)  from (iv):  1−k+(√(1+k^2 ))=(1−(1/m))(k/r)   ...(III)  putting I, II into III we get an. eqn.  for m.  ⇒m≈1.243756  ⇒r≈0.2694
eqn.ofOC:sayy=mxmxy=0eqn.ofAD:sauy=k(x1)kx+yk=0P(r,1r)Q(q,r)r=mr1+r1+m2(i)r=mqr1+m2(ii)r=kr+1rk1+k2(iii)r=kq+rk1+k2(iv)4unknowns:r,m,k,qmqr+mr1+r=0q=1mrkq+rk=kr+1rkk=12rqrk=12r1m2r(I)from(ii):r=1(m+1)r1+m2r=11+m+1+m2(II)from(iv):1k+1+k2=(11m)kr(III)puttingI,IIintoIIIwegetan.eqn.form.m1.243756r0.2694
Commented by mr W last updated on 18/Nov/21
Commented by ajfour last updated on 18/Nov/21
Excellent sir, thanks!  i m out of m town to take up a job,  hence my inconsistency..
Excellentsir,thanks!imoutofmtowntotakeupajob,hencemyinconsistency..
Commented by mr W last updated on 18/Nov/21
good luck with the new job sir!
goodluckwiththenewjobsir!
Commented by Tawa11 last updated on 28/Nov/21
Great sir
Greatsir

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