about to only mathematics

Explanation

The given is `ax^(2)+2hxy+by^(2)=1 " "...(i)`
Let the point P not lying on Eq. (i) be `(x_(1), y_(1))` , let `theta` be the inclination of line through P which intersect the given curve at Q and R.
Then, equation of line through P is
`(x-x_(1))/(costheta)=(y-y_(1))/(sintheta)=r`
`rArr x =x_(1)+r costheta, y = y_(1)+r sintheta`
For point Q and R, above point must lie on Eq. (i).
`rArr a (x_(1)+rcostheta)^(2)+2h(x_(1)+rcostheta)(y_(1)+rsintheta)`
`+b(y_(1)+rsintheta)^(2)=1`
`rArr(acos^(2)theta+2h sinthetacostheta+bsin^(2)theta)r^(2)`
`+2(ax_(1)costheta+hx_(1)sintheta+hy_(1)costheta+by_(1)sintheta)r`
`+(ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)-1)=0`
it is quadration r, giving two values of r as PQ and PR.
`therefore PQ*PR=(ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)-1)/(a cos^(2)theta+2hsinthetacostheta+bsin^(2)theta)`
Here, `ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)-1ne0,as (x_(1), y_(1))` does not lie on Eq. (i)
also, `acos^(2)theta+2hsinthetacostheta+bsin^(2)theta`
`=a+2hsintheta+(b-a)sin^(2)theta`
`=a+sintheta{2hcostheta+(b-a)sintheta}`
`=a+sintheta.sqrt(4h^(2)+(b-a)^(2)).(costhetasinphi+sinthetacosphi)`
where, `tantheta=(b-a)/(2h)`
`=a+sqrt(4h^(2)+(b-a)^(2))sintheta sin(theta+phi)`
which will be independent of `theta`, if
`4h^(2)+(b-a)^(2)=0`
`rArr h=0 and b=a`
`thereforeEq.(i) " reduces to " x^(2)+y^(2)=(1)/(a)`
which is equation of circle.