If a circle and the rectangular hyperbola xy=c2 meet in the four points whose parameters are p1,p2,p3 and p4, then the center of the circle through the points p1,p2 and p3 is
A
{c2(p1+p2+p3+1p1.p2.p3),c2(1p1+1p2+1p3+p1.p2.p3)}
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B
{c2(p1+p2+p3),c2(p1.p2.p3)}
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C
{c2(1p1.p2.p3),c2(1p1+1p2+1p3)}
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D
None of these
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Solution
The correct option is A {c2(p1+p2+p3+1p1.p2.p3),c2(1p1+1p2+1p3+p1.p2.p3)} Let the rectangular hyperbola be xy=c2 and the equation of the circle be x2+y2+2gx+2fy+k=0. Any point on the hyperbola is (cp,cp).If it lies on the circle, then c2p2+c2p2+2gcp+2fcp+k=0.
c2p4+2gcp3+kp2+2fcp+c2=0.
This is fourth-degree equation in p, which has four roots.
Hence the circle and the hyperbola intersect in four points.
If p1,p2,p3,p4 are the roots of this equation, thenp1+p2+p3+p4=2gcc2=2gccp1+cp2+cp3+cp4=−2gx1+x2+x3+x44=g2 ..(i)
Also 1p1+1p2+1p3+1p4=p1.p2.p3p1.p2.p3.p4=2fcc2=2fc
cp1+cp2+cp3+cp4=2f
y1+y2+y3+y44=f/2 ..(ii)
Also, p1.p2.p3.p4=c2c2=1 ..(iii)
Hence from (i),(ii) and (iii) we can clearly see that−g,−f=c2(p1+p2+p3+1p1.p2.p3),c2(1p1+1p2+1p3+p1.p2.p3)Hence, the correct option is (A).