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Question-124740




Question Number 124740 by ajfour last updated on 05/Dec/20
Commented by ajfour last updated on 05/Dec/20
Find eq. of circle passing through  intersection points of  curves  y=x^2 −1  and  y=(c/x) ;   (c < (2/(3(√3))))
Findeq.ofcirclepassingthroughintersectionpointsofcurvesy=x21andy=cx;(c<233)
Answered by ajfour last updated on 05/Dec/20
y=x^2 −1  y=(c/x)  let roots of  x^2 −1=(c/x)   be  p, q, r  p+q+r=0  pq+qr+rp=−1  pqr=c  (y−k)^2 +(x−h)^2 =R^( 2)   (x^2 −1−k)^2 +(x−h)^2 =R^( 2)   p(q^2 +r^2 −2−k)+p+2h=0  ⇒  p(p^2 −((2c)/p)−2−k)+p+2h=0  ⇒ 2h−(k+2)p+2p−c=0   ....(I)  ⇒ Σ  gives  ⇒    h=(c/2)  And considering (I) again  with  h=c  ⇒   k=0  ⇒  R^( 2)  = (x^2 −1)^2 +(x−(c/2))^2   ⇒  R^( 2) = 1+(c^2 /4)  Hence eq. of circle is    (x−(c/2))^2 +y^2 =1+(c^2 /4)  example:  For   c=(1/3)  , the curves are  y=x^2 −1  ,  y=(1/(3x)) ,  and the circle  (x−(1/6))^2 +y^2 =((37)/(36))   ★
y=x21y=cxletrootsofx21=cxbep,q,rp+q+r=0pq+qr+rp=1pqr=c(yk)2+(xh)2=R2(x21k)2+(xh)2=R2p(q2+r22k)+p+2h=0p(p22cp2k)+p+2h=02h(k+2)p+2pc=0.(I)Σgivesh=c2Andconsidering(I)againwithh=ck=0R2=(x21)2+(xc2)2R2=1+c24Henceeq.ofcircleis(xc2)2+y2=1+c24example:Forc=13,thecurvesarey=x21,y=13x,andthecircle(x16)2+y2=3736
Commented by ajfour last updated on 05/Dec/20
Commented by mr W last updated on 05/Dec/20
fantastic!
fantastic!
Commented by ajfour last updated on 06/Dec/20

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