Prove that the equation to the circle, which passes through the focus and touches the parabola $y^{2} = 4 a x$ at the point $(a t^{2}, 2 a t)$ , is
$x^{2} + y^{2} - a x (3 t^{2} + 1) - a y (3 t - t^{3}) + 3 a^{2} t^{2} = 0$ .
Prove also that the locus of its centre is the curve
$27 a y^{2} = (2 x - a) (x - 5 a)^{2}$ .

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Solution

$t y = x + a t^{2} x - t y + a t^{2} = 0$

Equation of circle is

${(x - a t^{2})}^{2} + {(y - 2 a t)}^{2} + λ (x - t y + a t^{2}) = 0...... (i)$

It passes through $(a, 0)$

${(a - a t^{2})}^{2} + {(0 - 2 a t)}^{2} + λ (a - t (0) + a t^{2}) = 0 λ (a + a t^{2}) = - (a^{2} + a^{2} t^{4} - 2 a^{2} t^{2} + 4 a^{2} t^{2}) λ = - \frac{a^{2} + a^{2} t^{4} + 2 a^{2} t^{2}}{a + a t^{2}} = - \frac{{(a + a t^{2})}^{2}}{a + a t^{2}} = - a (1 + t^{2})$

Substituting $λ$ in $(i)$ , we get

${(x - a t^{2})}^{2} + {(y - 2 a t)}^{2} - a (1 + t^{2}) (x - t y + a t^{2}) = 0 \Rightarrow x^{2} + y^{2} - a x (3 t^{2} + 1) - a y (3 t - t^{3}) + 3 a^{2} t^{2} = 0$

Let the centre of circle be $(h, k)$

$\Rightarrow h = \frac{a (3 t^{2} + 1)}{2}, k = \frac{a (3 t - t^{3})}{2} h = \frac{a (3 t^{2} + 1)}{2} 3 t^{2} + 1 = \frac{2 h}{a} t^{2} = \frac{2 h - a}{3 a} . . . . . . . . (i i) k = \frac{a (3 t - t^{3})}{2} k = \frac{a t (3 - t^{2})}{2} k^{2} = \frac{a^{2} t^{2} {(3 - t^{2})}^{2}}{4}$

Substituting $t^{2}$ from $(i i)$

$4 k^{2} = a^{2} (\frac{2 h - a}{3 a}) {(3 - \frac{2 h - a}{3 a})}^{2} 4 k^{2} = a^{2} (\frac{2 h - a}{3 a}) 4 {(\frac{- h + 5 a}{3 a})}^{2} 27 {a k}^{2} = (2 h - a) {(h - 5 a)}^{2}$

Generalising the equation

$27 {a y}^{2} = (2 x - a) {(x - 5 a)}^{2}$

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