Prove that, the normal to y2
= 12x at (3,
6) meets the parabola again in (27, −18) & circle on this normal
chord as diameter is x2
+ y2 − 30x + 12y − 27 = 0.

Equation of tangent to parabola y²=4ax at (x₁,y₁) is
y-y₁ = -y₁ (x-x₁)/2a
Equation of normal at (3,6) for above parabola will be
y- 6 = -6( x- 3)/(2*3)
On simplifying
x+ y =9
Now solve it with parabola by putting x=9-y in equation of parabola and find other root.
Now we have end points of diameter (3,6) and ( 27,-18)
equation of circle with (x₁,y₁) and (x₂,y₂) as diameter is
( x-x₁)(x-x₂) + (y-y₁)(y-y₂)=0
Put the values and get the required answer.