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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))))
$${Find}\:{eq}.\:{of}\:{circle}\:{passing}\:{through} \\ $$$${intersection}\:{points}\:{of}\:\:{curves} \\ $$$${y}={x}^{\mathrm{2}} −\mathrm{1}\:\:{and}\:\:{y}=\frac{{c}}{{x}}\:;\:\:\:\left({c}\:<\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$
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))   ★
$${y}={x}^{\mathrm{2}} −\mathrm{1} \\ $$$${y}=\frac{{c}}{{x}} \\ $$$${let}\:{roots}\:{of}\:\:{x}^{\mathrm{2}} −\mathrm{1}=\frac{{c}}{{x}}\:\:\:{be}\:\:{p},\:{q},\:{r} \\ $$$${p}+{q}+{r}=\mathrm{0} \\ $$$${pq}+{qr}+{rp}=−\mathrm{1} \\ $$$${pqr}={c} \\ $$$$\left({y}−{k}\right)^{\mathrm{2}} +\left({x}−{h}\right)^{\mathrm{2}} ={R}^{\:\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}−{k}\right)^{\mathrm{2}} +\left({x}−{h}\right)^{\mathrm{2}} ={R}^{\:\mathrm{2}} \\ $$$${p}\left({q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}−{k}\right)+{p}+\mathrm{2}{h}=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}\left({p}^{\mathrm{2}} −\frac{\mathrm{2}{c}}{{p}}−\mathrm{2}−{k}\right)+{p}+\mathrm{2}{h}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}{h}−\left({k}+\mathrm{2}\right){p}+\mathrm{2}{p}−{c}=\mathrm{0}\:\:\:….\left({I}\right) \\ $$$$\Rightarrow\:\Sigma\:\:{gives} \\ $$$$\Rightarrow\:\:\:\:{h}=\frac{{c}}{\mathrm{2}} \\ $$$${And}\:{considering}\:\left({I}\right)\:{again} \\ $$$${with}\:\:{h}={c}\:\:\Rightarrow\:\:\:{k}=\mathrm{0} \\ $$$$\Rightarrow\:\:{R}^{\:\mathrm{2}} \:=\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\left({x}−\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{R}^{\:\mathrm{2}} =\:\mathrm{1}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${Hence}\:{eq}.\:{of}\:{circle}\:{is} \\ $$$$\:\:\left({x}−\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${example}: \\ $$$${For}\:\:\:{c}=\frac{\mathrm{1}}{\mathrm{3}}\:\:,\:{the}\:{curves}\:{are} \\ $$$${y}={x}^{\mathrm{2}} −\mathrm{1}\:\:,\:\:{y}=\frac{\mathrm{1}}{\mathrm{3}{x}}\:,\:\:{and}\:{the}\:{circle} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{\mathrm{37}}{\mathrm{36}}\:\:\:\bigstar \\ $$
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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