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Explanation

The given curve is
`ax^(2)+2hxy+by^(2)=1` (1)
Let the point P not lying on curve (1) be `(x_(1),y_(1))` .
Let `theta` be the inclination of line through P which intersects the given curve at Q and R .
The equation of line through P is
`(x-x_(1))/(cos theta)=(y-y_(1))/(sin theta)=r`
or `x =r cos theta +x_(1)` and `y= r sin theta +y_(1)`
For point Q or R,the above point must lie on curve (1). So,
`a[x_(1)+r cos theta ]^(2)+2h[x_(1)+r cos theta ][y_(1)+r sin theta ] +b[y_(1)+rsin theta ]^(2)=1`
or `(a cos ^(2) theta +2h sin theta cos theta +b sin^(2) theta)r^92)+2(ax_(2) cos theta +hx_(1) sin theta +hx_(1)cos theta +by_(1) sin theta ) r +(ax_(1)^(2)+2h_(1)y_(1)+by_(1)^(2)-1)=0`
It is a quadratic in r giving two values of r for PQ and PR.
Therefore,
Product fo roots `= PQ. PR`
`=(ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)-1)/(acos ^(2)theta+2h sin theta cos theta +b sin ^(2) theta )`
Here, `ax_(1)^(2)+2hx_(1)y_(1)+by_(1)^(2)-1 cancel (=)0` as `(x_(1),y_(1))` does not lie on curve (1) . Also,
Denominator `=a cos^(2) theta +2h sin theta cos theta +b sin ^(2) theta`
`=a+2h sin theta cos theta +(b-a) sin^(2) theta`
`=a+h sin 2 theta +((b-a))/(2)(1-cos 2 theta)`
`=((a+b)/(2))+h sin 2 theta +((a-b)/(2))cos 2theta`
which is independent of `theta` if `h=0` , and `(a-b//2)=0, i.e., h=0` and `a=b`.
Hence, the given equation is a circle.