Prove that the equation to the circle passing through the points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ and the intersection of the tangents to the parabola at these points is
$x^{2} + y^{2} - a x [(t_{1} + t_{2})^{2} + 2] - a y (t_{1} + t_{2}) (1 - t_{1} t_{2}) + a^{2} t_{1} t_{2} (2 - t_{1} t_{2}) = 0 .$

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Solution

The point of intersection of tangents through $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ is $R (a t_{1} t_{2}, a (t_{1} + t_{2}))$
Equation of chord $P Q$ is
$(t_{1} + t_{2}) y = 2 x + 2 a t_{1} t_{2} 2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0$
Equation of circle through end points of chord $P Q$ is
$(x - a t_{1}^{2}) (x - a t_{2}^{2}) + (y - 2 a t_{1}) (y - 2 a t_{2}) + λ (2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2}) = 0..... (i)$
It passes through $R$
$(a t_{1} t_{2} - a t_{1}^{2}) (a t_{1} t_{2} - a t_{2}^{2}) + (a (t_{1} + t_{2}) - 2 a t_{1}) (a (t_{1} + t_{2}) - 2 a t_{2}) + λ (2 a t_{1} t_{2} - (t_{1} + t_{2}) a (t_{1} + t_{2}) + 2 a t_{1} t_{2}) = 0 - a t_{1} t_{2} {(t_{1} - t_{2})}^{2} - a {(t_{1} - t_{2})}^{2} - λ {(t_{1} - t_{2})}^{2} = 0 \Rightarrow λ = - a (t_{1} t_{2} + 1)$
Simplifying $(i)$
$x^{2} + y^{2} - x (a t_{1}^{2} + a t_{2}^{2} - 2 λ) - y (2 a t_{1} + 2 a t_{2} + (t_{1} + t_{2}) λ) + a^{2} t_{1}^{2} t_{2}^{2} + 4 a^{2} t_{1} t_{2} + 2 a t_{1} t_{2} λ = 0$
Substituting $λ$
$x^{2} + y^{2} - x {a t_{1}^{2} + a t_{2}^{2} - 2 (- a (t_{1} t_{2} + 1)} - y {2 a t_{1} + 2 a t_{2} + (t_{1} + t_{2}) (- a (t_{1} t_{2} + 1))} + a^{2} t_{1}^{2} t_{2}^{2} + 4 a^{2} t_{1} t_{2} + 2 a t_{1} t_{2} (- a (t_{1} t_{2} + 1)) = 0 x^{2} + y^{2} - x (a t_{1}^{2} + a t_{2}^{2} + 2 a t_{1} t_{2} + 2 a) - a y (t_{1} - t_{2} t_{1}^{2} + t_{2} - t_{1} t_{2}^{2}) + a^{2} (t_{1}^{2} t_{2}^{2} + 4 t_{1} t_{2} - 2 t_{1}^{2} t_{2}^{2} - 2 t_{1} t_{2}) = 0 x^{2} + y^{2} - a x {{(t_{1} + t_{2})}^{2} + 2} - a y (t_{1} + t_{2}) (1 - t_{1} t_{2}) + a^{2} t_{1} t_{2} (2 - t_{1} t_{2}) = 0$
Hence proved.

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