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Question Number 122456 by ajfour last updated on 17/Nov/20

Commented by ajfour last updated on 17/Nov/20

y=1+(c/x) (0<c<(2/(3(√3)))) y=x^2 both these curves intersect at P, Q, R as shown; find eq. of circle through these three points without finding the coordinates of the points of intersection.

$${y}=\mathrm{1}+\frac{{c}}{{x}}\:\:\:\:\:\:\:\:\:\left(\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$$${y}={x}^{\mathrm{2}} \\ $$$${both}\:{these}\:{curves}\:{intersect}\:{at} \\ $$$${P},\:{Q},\:{R}\:\:{as}\:{shown};\:\:{find}\:{eq}.\:{of} \\ $$$${circle}\:{through}\:{these}\:{three}\:{points} \\ $$$${without}\:{finding}\:{the}\:{coordinates} \\ $$$${of}\:{the}\:{points}\:\:{of}\:{intersection}. \\ $$

Answered by ajfour last updated on 21/Nov/20

y=x^2 ; y=1+(c/x) let x coordinates of intersection points be p, q, r. They are roots of eq. x^3 −x−c=0 ⇒ p+q+r=0 pq+qr+rp=−1 pqr=c Eq. of line ⊥ to QR and passing through mid point of QR is y−(((q^2 +r^2 )/2))=−(((x−((q+r)/2)))/(q+r)) but q+r=−p and q^2 +r^2 = p^2 −((2c)/p) , hence 2y−(p^2 −((2c)/p))=((2x+p)/p) As, 1+(c/p)=p^2 ⇒ p^2 −((2c)/p)+1=2−(c/p) ⇒ 2y=((2x−c)/p)+2 ⇒ p(y−1)=x−(c/2) .....(I) in a similar way eq. of line ⊥ to chord PQ of circle should be r(y−1)=x−(c/2) .....(II) and eq. of diameter ⊥ to chord PR is q(y−1)=x−(c/2) ....(III) Let center be (h,k) Adding (I), (II), (III) (p+q+r)(k−1)=3h−((3c)/2) ⇒ h=(c/2) & k=1 radius of circle 𝛒 is found from 3ρ^2 =Σ{((c/2)−p)^2 +(1−p^2 )^2 } =((3c^2 )/4)−Σp^2 −3cΣp+3−2Σp^2 +3Σp^4 but Σp=p+q+r=0 hence 3ρ^2 =((3c^2 )/4)−3Σp^2 +3−+3Σ(p^2 +cp) ⇒ 3𝛒^2 =((3c^2 )/4)+3 𝛒=(c^2 /4)+1 .....(II) ____________________________ Hence eq. of circle is (x−(c/2))^2 +(y−1)^2 = (c^2 /4)+1 ____________________________

$${y}={x}^{\mathrm{2}} \:\:\:;\:\:{y}=\mathrm{1}+\frac{{c}}{{x}} \\ $$$${let}\:{x}\:{coordinates}\:{of}\:{intersection} \\ $$$${points}\:{be}\:\:\boldsymbol{{p}},\:\boldsymbol{{q}},\:\boldsymbol{{r}}. \\ $$$$\:{They}\:{are}\:{roots}\:{of}\:{eq}.\:\:\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{x}}−\boldsymbol{{c}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{p}}+\boldsymbol{{q}}+\boldsymbol{{r}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{pq}}+\boldsymbol{{qr}}+\boldsymbol{{rp}}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{pqr}}=\boldsymbol{{c}} \\ $$$${Eq}.\:{of}\:\:{line}\:\bot\:{to}\:{QR}\:{and}\:{passing} \\ $$$${through}\:{mid}\:{point}\:{of}\:{QR}\:{is} \\ $$$$\:\:{y}−\left(\frac{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}}\right)=−\frac{\left({x}−\frac{{q}+{r}}{\mathrm{2}}\right)}{{q}+{r}}\: \\ $$$${but}\:\:\:{q}+{r}=−{p}\:\: \\ $$$${and}\:\:{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\:{p}^{\mathrm{2}} −\frac{\mathrm{2}{c}}{{p}}\:\:\:\:\:,\:{hence}\: \\ $$$$\:\:\mathrm{2}{y}−\left({p}^{\mathrm{2}} −\frac{\mathrm{2}{c}}{{p}}\right)=\frac{\mathrm{2}{x}+{p}}{{p}} \\ $$$${As},\:\:\:\mathrm{1}+\frac{{c}}{{p}}={p}^{\mathrm{2}} \:\Rightarrow\:\:{p}^{\mathrm{2}} −\frac{\mathrm{2}{c}}{{p}}+\mathrm{1}=\mathrm{2}−\frac{{c}}{{p}} \\ $$$$\Rightarrow\:\:\mathrm{2}{y}=\frac{\mathrm{2}{x}−{c}}{{p}}+\mathrm{2} \\ $$$$\Rightarrow\:\:{p}\left({y}−\mathrm{1}\right)={x}−\frac{{c}}{\mathrm{2}}\:\:\:\:\:\:.....\left({I}\right) \\ $$$${in}\:{a}\:{similar}\:{way}\:{eq}.\:{of}\:{line}\:\bot\:{to} \\ $$$${chord}\:{PQ}\:{of}\:{circle}\:{should}\:{be} \\ $$$$\:\:\:\:\:\:\:{r}\left({y}−\mathrm{1}\right)={x}−\frac{{c}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:.....\left({II}\right) \\ $$$${and}\:{eq}.\:{of}\:\:{diameter}\:\:\bot\:{to}\:{chord} \\ $$$${PR}\:{is}\:\:\:\:\:\:{q}\left({y}−\mathrm{1}\right)={x}−\frac{{c}}{\mathrm{2}}\:\:\:\:\:\:....\left({III}\right) \\ $$$${Let}\:\:\:{center}\:{be}\:\:\:\left({h},{k}\right) \\ $$$${Adding}\:\left({I}\right),\:\left({II}\right),\:\left({III}\right) \\ $$$$\:\:\:\:\:\left({p}+{q}+{r}\right)\left({k}−\mathrm{1}\right)=\mathrm{3}{h}−\frac{\mathrm{3}{c}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\boldsymbol{{h}}=\frac{\boldsymbol{{c}}}{\mathrm{2}}\:\:\:\:\&\:\:\:\boldsymbol{{k}}=\mathrm{1} \\ $$$${radius}\:{of}\:{circle}\:\boldsymbol{\rho}\:{is}\:{found}\:{from} \\ $$$$\:\:\:\:\mathrm{3}\rho^{\mathrm{2}} =\Sigma\left\{\left(\frac{{c}}{\mathrm{2}}−{p}\right)^{\mathrm{2}} +\left(\mathrm{1}−{p}^{\mathrm{2}} \right)^{\mathrm{2}} \right\} \\ $$$$\:\:=\frac{\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{4}}−\Sigma{p}^{\mathrm{2}} −\mathrm{3}{c}\Sigma{p}+\mathrm{3}−\mathrm{2}\Sigma{p}^{\mathrm{2}} +\mathrm{3}\Sigma{p}^{\mathrm{4}} \\ $$$$\:{but}\:\:\Sigma{p}={p}+{q}+{r}=\mathrm{0}\:\:{hence} \\ $$$$\:\:\mathrm{3}\rho^{\mathrm{2}} =\frac{\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}\Sigma{p}^{\mathrm{2}} +\mathrm{3}−+\mathrm{3}\Sigma\left({p}^{\mathrm{2}} +{cp}\right) \\ $$$$\Rightarrow\:\:\mathrm{3}\boldsymbol{\rho}^{\mathrm{2}} =\frac{\mathrm{3}\boldsymbol{{c}}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3} \\ $$$$\:\:\:\:\boldsymbol{\rho}=\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}\:\:\:\:\:.....\left(\boldsymbol{{II}}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\:\:{Hence}\:{eq}.\:{of}\:{circle}\:{is} \\ $$$$\:\:\:\left({x}−\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\:\frac{{c}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$ \\ $$

Commented by mr W last updated on 21/Nov/20

great!

$${great}! \\ $$

Commented by ajfour last updated on 21/Nov/20

thanks sir!

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