Prove that the equation to the circle, which passes through the focus and touches the parabola y 2 = 4 a x a > 0 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 . Also prove that the locus of its centre is the curve 27 a y 2 = 2 x - a x - 5 a 2 .

If the circle touches the parabola y 2 = 4 a x at ( a t 2 , 2 a t ) , then they must have a common tangent at that point, and hence a common normal. The centre of the circle must lie on that normal. Let ( h , k ) be the coordinates of the centre and r be the radius of the circle. Hence, its equation can be written as x 2 + y 2 - 2 h x - 2 k y + c = 0 . . . ( i ) The equation to the normal at ( a t 2 , 2 a t ) is y = - t x + 2 a t + a t 3 . . . ( ii ) As the centre ( h , k ) lies on the normal ( ii ) . Hence, k = - t h + 2 a t + a t 3 . . . ( iii ) Focus of the parabola is ( a , 0 ) . As the circle passes through ( a , 0 ) and ( a t 2 , 2 a t ) distance of these points from the centre ( h , k ) must be equal to the radius. ⇒ r 2 = h - a 2 + k 2 = h - a t 2 2 + k - 2 a t 2 ⇒ - 2 a h + a 2 = - 2 a h t 2 - 4 a k t + a 2 t 4 + 4 a 2 t 2 Dividing by a ⇒ 4 k t = 2 h ( 1 - t 2 ) + a t 4 + 4 a t 2 - a . . . ( iv ) Solving ( iii ) and ( iv ) , 4 t ( - t h + 2 a t + a t 3 ) = 2 h ( 1 - t 2 ) + a t 4 + 4 a t 2 - a ⇒ - 4 h t 2 + 8 a t 2 + 4 a t 4 = a t 4 - 2 h t 2 + 4 a t 2 + 2 h - a ⇒ 3 a t 4 - 2 h t 2 + 4 a t 2 - 2 h + a = 0 ⇒ a ( 3 t 4 + 4 t 2 + 1 ) = 2 h ( 1 + t 2 ) ⇒ ( 3 t 4 + 3 t 2 + t 2 + 1 ) = 2 h ( 1 + t 2 ) ⇒ a ( 3 t 2 + 1 ) ( 1 + t 2 ) = 2 h ( 1 + t 2 ) 2 h = a ( 3 t 2 + 1 ) . . . ( v ) and 2 k = a ( 3 t - t 3 ) . . . ( vi ) Putting the values of h and k in ( i ) , Equation of the circle is x 2 + y 2 - a x ( 3 t 2 + 1 ) - a y ( 3 t - t 3 ) y + c = 0 . . . ( vii ) As ( vii ) passes through the focus ( a , 0 ) , the coordinates will satisfy it. Hence, a 2 - a 2 ( 3 t 2 + 1 ) + c = 0 ⇒ c = 3 a 2 t 2 . Putting the values of c in ( vii ) , the circle is x 2 + y 2 - a x 3 t 2 + 1 - a y 3 t - t 3 + 3 a 2 t 2 = 0 Hence proved. Now, to get the locus of centre, we have to eliminate t from ( v ) and ( vi ) . Multiplying ( v ) by t and ( vi ) by 3 and adding them, 2 h t = a t ( 3 t 2 + 1 ) ⇒ 6 k = 3 a ( 3 t - t 3 ) ⇒ 2 t h + 6 k = 10 a t t = 3 k 5 a - h Using the equation ( v ) , 2 h = a ( 3 t 2 + 1 ) ⇒ 2 h - a = 3 a 3 k 5 a - h 2 ⇒ 2 h - a 5 a - h 2 = 27 a k 2 Simplifying and generalizing, the locus of centre ( h , k ) is 2 x - a x - 5 a 2 = 27 a y 2