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

Commented by ajfour last updated on 17/Nov/20

$y = 1 + \frac{c}{x} (0 < c < \frac{2}{3 \sqrt{3}})$ $y = x^{2}$ $b o t h t h e s e c u r v e s i n t e r s e c t a t$ $P, Q, R a s s h o w n; f i n d e q . o f$ $c i r c l e t h r o u g h t h e s e t h r e e p o i n t s$ $w i t h o u t f i n d i n g t h e c o o r d i n a t e s$ $o f t h e p o i n t s o f i n t e r s e c t i o n .$
	Answered by ajfour last updated on 21/Nov/20
	$y=x^2 ; y=1+(c/x) let x coordinates of intersection points be p, q, r. They are roots of eq. x^3 −x−c=0 ⇒ p+q+r=0 pq+qr+rp=−1 pqr=c Eq. of line ⊥ to QR and passing through mid point of QR is y−(((q^2 +r^2 )/2))=−(((x−((q+r)/2)))/(q+r)) but q+r=−p and q^2 +r^2 = p^2 −((2c)/p) , hence 2y−(p^2 −((2c)/p))=((2x+p)/p) As, 1+(c/p)=p^2 ⇒ p^2 −((2c)/p)+1=2−(c/p) ⇒ 2y=((2x−c)/p)+2 ⇒ p(y−1)=x−(c/2) .....(I) in a similar way eq. of line ⊥ to chord PQ of circle should be r(y−1)=x−(c/2) .....(II) and eq. of diameter ⊥ to chord PR is q(y−1)=x−(c/2) ....(III) Let center be (h,k) Adding (I), (II), (III) (p+q+r)(k−1)=3h−((3c)/2) ⇒ h=(c/2) & k=1 radius of circle 𝛒 is found from 3ρ^2 =Σ{((c/2)−p)^2 +(1−p^2 )^2 } =((3c^2 )/4)−Σp^2 −3cΣp+3−2Σp^2 +3Σp^4 but Σp=p+q+r=0 hence 3ρ^2 =((3c^2 )/4)−3Σp^2 +3−+3Σ(p^2 +cp) ⇒ 3𝛒^2 =((3c^2 )/4)+3 𝛒=(c^2 /4)+1 .....(II) ____________________________ Hence eq. of circle is (x−(c/2))^2 +(y−1)^2 = (c^2 /4)+1 ____________________________$
	$y = x^{2}; y = 1 + \frac{c}{x}$ $l e t x c o o r d i n a t e s o f i n t e r s e c t i o n$ $p o i n t s b e p, q, r .$ $T h e y a r e r o o t s o f e q . x^{3} - x - c = 0$ $\Rightarrow p + q + r = 0$ $p q + q r + r p = - 1$ $p q r = c$ $E q . o f l i n e ⊥ t o Q R a n d p a s s i n g$ $t h r o u g h m i d p o i n t o f Q R i s$ $y - (\frac{q^{2} + r^{2}}{2}) = - \frac{(x - \frac{q + r}{2})}{q + r}$ $b u t q + r = - p$ $a n d q^{2} + r^{2} = p^{2} - \frac{2 c}{p}, h e n c e$ $2 y - (p^{2} - \frac{2 c}{p}) = \frac{2 x + p}{p}$ $A s, 1 + \frac{c}{p} = p^{2} \Rightarrow p^{2} - \frac{2 c}{p} + 1 = 2 - \frac{c}{p}$ $\Rightarrow 2 y = \frac{2 x - c}{p} + 2$ $\Rightarrow p (y - 1) = x - \frac{c}{2} . . . . . (I)$ $i n a s i m i l a r w a y e q . o f l i n e ⊥ t o$ $c h o r d P Q o f c i r c l e s h o u l d b e$ $r (y - 1) = x - \frac{c}{2} . . . . . (I I)$ $a n d e q . o f d i a m e t e r ⊥ t o c h o r d$ $P R i s q (y - 1) = x - \frac{c}{2} . . . . (I I I)$ $L e t c e n t e r b e (h, k)$ $A d d i n g (I), (I I), (I I I)$ $(p + q + r) (k - 1) = 3 h - \frac{3 c}{2}$ $\Rightarrow h = \frac{c}{2} & k = 1$ $r a d i u s o f c i r c l e ρ i s f o u n d f r o m$ $3 ρ^{2} = Σ {{(\frac{c}{2} - p)}^{2} + {(1 - p^{2})}^{2}}$ $= \frac{3 c^{2}}{4} - Σ p^{2} - 3 c Σ p + 3 - 2 Σ p^{2} + 3 Σ p^{4}$ $b u t Σ p = p + q + r = 0 h e n c e$ $3 ρ^{2} = \frac{3 c^{2}}{4} - 3 Σ p^{2} + 3 - + 3 Σ (p^{2} + c p)$ $\Rightarrow 3 ρ^{2} = \frac{3 c^{2}}{4} + 3$ $ρ = \frac{c^{2}}{4} + 1 . . . . . (I I)$ $____________________________$ $H e n c e e q . o f c i r c l e i s$ ${(x - \frac{c}{2})}^{2} + {(y - 1)}^{2} = \frac{c^{2}}{4} + 1$ $____________________________$
	Commented by mr W last updated on 21/Nov/20

	$g r e a t!$
	Commented by ajfour last updated on 21/Nov/20
	thanks sir!
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