# A, B are fixed points; AP and BQ are parallel chords of a variable circle such thatAP/BQ is a constant. Prove that the locus of P is a circle.

Yuvraj Singh
2 years ago
Ans:
We takeOas origin and x-axis alongBOA. Let the coordinates ofAandBbe(a,0)and(−b,0). Let the fixed parallel linesAPandBQbe taken parallel to y-axis so thatBOAis perpendicular to both.
If∠POA=θthen as∠POQ=90o
We have∠QOB=90o−θ
Coordinates ofPare(a,atanθ)and those ofQare
(−b,bcotθ)
Equation to linePQis
y−atanθ=−(b+a)bcotθ−atanθ​(x−a)
orx(bcotθ−atanθ)+y(a+b)=ab(cotθ+tanθ)
or changing intosinθandcosθit becomes
x(bcos2θ−asin2θ)+y(a+b)sinθcosθ=ab
Multiplying by2, it becomes
x(2bcos2θ−2asin2θ)+y(a+b)sin2θ=2ab....(1)
Also equation to perpendicularORonPQis
x(a+b)sin2θ−y(2bcos2θ−2asin2θ)=0....(2)
But(2bcos2θ−2asin2θ)=b(1+cos2θ)−a(1+cos2θ)=(a+b)1+cos2θ−(a−b)
Hence the two lines (1) and (2) can be written as
x(a+b)cos2θ+y(a+b)sin2θ=2ab+(a−b)x.....(3)
andx(a+b)sin2θ−y(a+b)cos2θ=−(a−b)y.....(4)
In order to find the locus of point of intersectionR, we have to eliminateθfor which we square and add the equations (3) and (4)
∴x2(a+b)2+y2(a+b)2=4a2b2+4ab(a−b)x+(a−b)2(x2+y2)
or(x2+y2)((a+b)2−(a−b)2)=4ab(ab+(a−b)x)
x2−(a−b)x−ab+y2=0or(x−a)(y+b)+y2=0
above represents the equation of a circle onABas diameter.